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ELEMENTARY  CALCULUS 


BY 
FREDERICK  S.  WOODS 

AND 

FREDERICK  H.  BAILEY 

PROFESSORS  OF  MATHEMATICS  IN  THE  MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY 


GINN  AND  COMPANY 

BOSTON     .    NEW   YORK     •     CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •     COLUMBUS     •     SAN   FRANCISCO 


COPYRIGHT,    1922,    BY    FREDERICK    S.    WOODS 

AND    FREDERICK    H.    BAILEY 

ALL  RIGHTS  RESERVED 


gfte  fltftcngum   gre« 

GINN  AND  COMPANY  •  PRO- 
PRIETORS  •  BOSTON  •  U.S.A. 


"^l^ 


PREFACE 

This  book  is  adapted  to  the  use  of  students  in  the  first  year 
in  technical  school  or  college,  and  is  based  upon  the  experience 
of  the  authors  in  teaching  calculus  to  students  in  the  Massa- 
chusetts Institute  of  Technology  immediately  upon  entrance. 
It  is  accordingly  assumed  that  the  student  has  had  college- 
entrance  algebra,  including  graphs,  and  an  elementary  course 
in  trigonometry,  but  that  he  has  not  studied  analytic  geometry. 

The  first  three  chapters  form  an  introductory  course  in 
which  the  fundamental  ideas  of  the  calculus  are  introduced, 
including  derivative,  differential,  and  the  definite  integral,  but 
the  formal  work  is  restricted  to  that  involving  only  the  poly- 
nomial. These  chapters  alone  are  well  fitted  for  a  short  course 
of  about  a  term. 

The  definition  of  the  derivative  is  obtained  through  the 
concept  of  speed,  using  familiar  illustrations,  and  the  idea 
of  a  derivative  as  measuring  the  rate  of  change  of  related 
quantities  is  emphasized.  The  slope  of  a  curve  is  introduced 
later.  This  is  designed  to  prevent  the  student  from  acquiring 
the  notion  that  the  derivative  is  fundamentally  a  geometric 
concept.  For  the  same  reason,  problems  from  mechanics  are 
prominent  throughout  the  book. 

With  Chapter  IV  a  more  formal  development  of  the  subject 
begins,  and  certain  portions  of  analytic  geometry  are  introduced 
as  needed.  These  include,  among  other  things,  the  straight  line, 
the  conic  sections,  the  cycloid,  and  polar  coordinates. 

The  book  contains  a  large  number  of  well-graded  exercises  for 
the  student.  Drill  exercises  are  placed  at  the  end  of  most  sec- 
tions, and  a  miscellaneous  set  of  exercises,  for  review  or  further 
work,  is  found  at  the  end  of  each  chapter  except  the  first. 

iii 

ivi305109 


IV  PREFACE 

Throughout  the  book,  the  authors  believe,  the  matter  is  pre- 
sented in  a  manner  which  is  well  within  the  capacity  of  a  first- 
year  student  to  understand.  They  liave  endeavored  to  teach 
the  calculus  from  a  common-sense  standpoint  as  a  very  useful 
tool.  They  have  used  as  much  mathematical  rigor  as  the 
student  is  able  to  understand,  but  have  refrained  from  raising 
the  more  difficult  questions  which  the  student  in  his  first 
course  is  able  neither  to  appreciate  nor  to  master. 

Students  who  have  completed  this  text  and  wish  to  continue 
their  study  of  mathematics  may  next  take  a  brief  course  in 
differential  equations  and  then  a  course  in  advanced  calculus, 
or  they  may  take  a  course  in  advanced  calculus  which  includes 
differential  equations.  It  would  also  be  desirable  for  such  stu- 
dents to  have  a  brief  course  in  analytic  geometry,  which  may 
either  follow  this  text  directly  or  come  later. 

This  arrangement  of  work  the  authors  consider  preferable  to 
the  one  —  for  a  long  time  common  in  American  colleges  —  by 
which  courses  in  higher  algebra  and  analytic  geometry  precede 
the  calculus.  However,  the  teacher  who  prefers  to  follow  the 
older  arrangement  will  find  this  text  adapted  to  such  a  program. 

F.  S.  WOODS 
F.  H.  BAILEY 


CONTENTS 


CHAPTER  I.     RATES 

SECTION  PAGE 

1.  Limits 1 

2.  Average  speed 3 

3.  True  speed 5 

4.  Algebraic  method 8 

5.  Acceleration 9 

6.  Rate  of  change 11 

CHAPTER  II.     DIFFERENTIATION 

7.  The  derivative 15 

8.  Differentiation  of  a  polynomial 18 

9.  Sign  of  the  derivative 20 

10.  Velocity  and  acceleration  (continued) 21 

11.  Rate  of  change  (continued) 24 

12.  Graphs 27 

13.  Real  roots  of  an  equation 30 

14.  Slope  of  a  straight  line  . 31 

15.  Slope  of  a  curve 36 

16.  The  second  derivative 39 

17.  Maxima  and  minima 41 

18.  Integration 44 

19.  Area 47 

20.  Differentials 50 

21.  Approximations 53 

General  exercises ,     ...  55 

CHAPTER  III.     SUMMATION 

22.  Area  by  summation 60 

23.  The  definite  integral 62 

24.  The  general  summation  problem 66 

25.  Pressure 68 

26.  Volume 71 

General  exercises 76 


vi  CONTENTS 

CHAPTER  IV.     ALGEBRAIC  FUNCTIONS 

SECTION  PAGE 

27.  Distance  between  two  points 79 

28.  Circle 79 

29.  Parabola 81 

30.  Parabolic  segment ' 83 

31.  Ellipse 85 

32.  Hyperbola  . 87 

33.  Other  curves 91 

34.  Theorems  on  limits 93 

35.  Theorems  on  derivatives 94 

36.  Formulas 101 

37.  Differentiation  of  implicit  functions 102 

38.  Tangent  line 104 

39.  The  differentials  dx,  dy,  ds 106 

40.  Motion  in  a  curve 107 

41.  Related  velocities  and  rates Ill 

General  exercises 113 


CHAPTER  V.     TRIGONOMETRIC  FUNCTIONS 

42.  Circular  measure 119 

43.  Graphs  of  trigonometric  functions 121 

44.  Differentiation  of  trigonometric  functions 124 

45.  Simple  harmonic  motion 127 

46.  Graphs  of  inverse  trigonometric  functions 130 

47.  Differentiation  of  inverse  trigonometric  functions 131 

48.  Angular  velocity 135 

49.  The  cycloid *" 137 

50.  Curvature 139 

51.  Polar  coordinates 142 

52.  The  differentials  dr,  dd,  ds,  in  polar  coordinates 146 

General  exercises 149 


CHAPTER  VI.     EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

53.  The  exponential  function 154 

54.  The  logarithm 154 

55.  Certain  empirical  equations 159 

56.  Differentiation 163 

57.  The  compound-interest  law 166 

General  exercises 168 


CONTENTS  vii 
CHAPTER  VII.     SERIES 

SECTION  PAGE 

58.  Power  series 172 

59.  Maclaurin's  series 173 

60.  Taylor's  series 177 

General  exercises 179 

CHAPTER  VIII.     PARTIAL  DIFFERENTIATION 

61.  Partial  differentiation 181 

62.  Higher  partial  derivatives 184 

63.  Total  differential  of  a  function  of  two  variables 185 

64.  Rate  of  change 189 

General  exercises 191 

CHAPTER  IX.     INTEGRATION 

65.  Introduction 194 

66.  Integral  of  m« 195 

67-68.  Other  algebraic  integrands 199 

69.  Integrals  of  trigonometric  functions 205 

70.  Integrals  of  exponential  functions 207 

71-72.  Substitutions 208 

73-74.  Integration  by  parts 212 

75.  Integration  of  rational  fractions 216 

76.  Table  of  integrals 217 

General  exercises 220 

CHAPTER  X.     APPLICATIONS 

77.  Review  problems 225 

78.  Infinite  limits,  or  integrand 229 

79.  Area  in  polar  coordinates 230 

80.  Mean  value  of  a  function 233 

81.  Length  of  a  plane  curve 235 

82.  Work 237 

General  exercises 239 

CHAPTER  XI.  REPEATED  INTEGRATION 

83.  Double  integrals 244 

84.  Area  as  a  double  integral 246 

85.  Center  of  gravity 249 


viii  CONTENTS 

SECTION  PAGE 

86.  Center  of  gravity  of  a  composite  area 255 

87.  Theorems 257 

88.  Moment  of  inertia 260 

89.  Moments  of  inertia  about  parallel  axes 266 

90.  Space  coordinates 269 

91.  Certain  surfaces ...  271 

92.  Volume 277 

93.  Center  of  gravity  of  a  solid 282 

94.  Moment  of  inertia  of  a  solid 283 

General  exercises 286 

ANSWERS 291 

INDEX 315 


ELEMENTARY  CALCULUS 


CHAPTER  I 
RATES 

1.  Limits.  Since  the  calculus  is  based  upon  the  idea  of  a 
limit  it  is  necessary  to  have  a  clear  understanding  of  the  word. 
Two  examples  already  familiar  to  the  student  will  be  sufficient. 

In  finding  the  area  of  a  circle  in  plane  geometry  it  is  usual 
to  begin  by  inscribing  a  regular  polygon  in  the  circle.  The  area 
of  the  polygon  differs  from  that  of  the  circle  by  a  certain 
amount.  As  the  number  of  sides  of  the  polygon  is  increased, 
this  difference  becomes  less  and  less.  Moreover,  if  we  take  any 
small  number  e,  we  can  find  an  inscribed  polygon  whose  area 
differs  from  that  of  the  circle  by  less  than  e ;  and  if  one  such 
polygon  has  been  found,  any  polygon  with  a  larger  number  of 
sides  will  still  differ  in  area  from  the  circle  by  less  than  e.  The 
area  of  the  circle  is  said  to  be  the  limit  of  the  area  of  the 
inscribed  polygon. 

As  another  example  of  a  limit  consider  the  geometric  progres- 
sion with  an  unlimited  number  of  terms 

i  +  J  +  i  +  U--- 

The  sum  of  the  first  two  terms  of  this  series  is  1|,  the  sum 
of  the  first  three  terms  is  1|,  the  sum  of  the  first  four  terms 
is  1|,  and  so  on.  It  may  be  found  by  trial  and  is  proved  in 
the  algebras  that  the  sum  of  the  terms  becomes  more  nearly 
equal  to  2  as  the  number  of  terms  which  are  taken  becomes 
greater.  Moreover,  it  may  be  shown  that  if  any  small  number 
e  is  assumed,  it  is  possible  to  take  a  number  of  terms  n  so  that 
the  sum  of  these  terms  differs  from  2  by  less  than  e.  If  a  value 
of  n  has  thus  been  found,  then  the  sum  of  a  number  of  terms 

1 


2  RATES 

greater  than  n  will  still  differ  from  2  by  less  than  e.  The 
number  2  is  said  to  be  the  limit  of  the  sum  of  the  first  n  terms 
of  the  series. 

In  each  of  these  two  examples  there  is  a  certain  variable  — 
namely,  the  area  of  the  inscribed  polygon  of  n  sides  in  one  case 
and  the  sum  of  the  first  n  terms  of  the  series  in  the  other  case — 
and  a  certain  constant,  the  area  of  the  circle  and  the  number  2 
respectively.  In  each  case  the  difference  between  the  constant 
and  the  variable  may  be  made  less  than  any  small  number  e  by 
taking  n  sufficiently  large,  and  this  difference  then  continues 
to  be  less  than  e  for  any  larger  value  of  n. 

This  is  the  essential  property  of  a  limit,  which  may  be  defined 
as  follows: 

A  constant  A  is  said  to  he  the  limit  of  a  variable  Xif  as  the  vari- 
able changes  its  value  according  to  some  law,  the  difference  between 
the  variable  and  the  constant  becomes  and  remains  less  than  any 
small  quantity  which  may  be  assigned. 

The  definition  does  not  say  that  the  variable  never  reaches  its 
limit.  In  most  cases  in  this  book,  however,  the  variable  fails  to 
do  so,  as  in  the  two  examples  already  given.  For  the  polygon  is 
never  exactly  a  circle,  nor  is  the  sum  of  the  terms  of  the  series 
exactly  2.  Examples  may  be  given,  however,  of  a  variable's 
becoming  equal  to  its  limit,  as  in  the  case  of  a  swinging  pendulum 
finally  coming  to  rest.  But  the  fact  that  a  variable  may  never 
reach  its  limit  does  not  make  the  limit  inexact.  There  is  nothing 
inexact  about  the  area  of  a  circle  or  about  the  number  2. 

The  student  should  notice  the  significance  of  the  word 
"remains"  in  the  definition.  If  a  railroad  train  approaches  a 
station,  the  difference  between  the  position  of  the  train  and 
a  point  on  the  track  opposite  the  station  becomes  less  than  any 
number  which  may  be  named ;  but  if  the  train  keeps  on  by  the 
station,  that  difference  does  not  remain  small.  Hence  there  is 
no  limit  approached  in  this  case. 

If  X  is  a  variable  and  A  a  constant  which  X  approaches  as  a 
limit,  it  follows  from  the  definition  that  we  may  write 

X=A^e,  (1) 


SPEED  3 

where  e  is  a  quantity  (not  necessarily  positive)  which  may  be 
made,  and  then  will  remain,  as  small  as  we  please. 

Conversely,  if  as  the  result  of  any  reasoning  we  arrive  at  a 
formula  of  the  form  (1)  where  X  is  a  variable  and  A  a  constant, 
and  if  we  see  that  we  can  make  e  as  small  as  we  please  and 
that  it  will  then  remain  just  as  small  or  smaller  as  X  varies, 
we  can  say  that  A  is  the  limit  of  X.  It  is  in  this  way  that  we 
shall  determine  limits  in  the  following  pages. 

2.  Average  speed.  Let  us  suppose  a  body  (for  example,  an 
automobile)  moving  from  a  point  ^  to  a  point  B  (Fig.  1),  a 
distance  of  100  mi.  If  the  automobile  takes  5  hr.  for  the  trip, 
we  are  accustomed  to  say  that  it  has  traveled  at  the  rate  of 
20  mi.  an  hour.   Everybody  knows  .  FOB 

that  this  does  not  mean  that  the  '  ""*     '  ' 

automobile  went  exactly  20  mi.  in 

each  hour  of  the  trip,  exactly  10  mi.  in  each  half  hour,  exactly 
6  mi.  in  each  quarter  hour,  and  so  on.  Probably  no  automobile 
ever  ran  in  such  a  way  as  that.  The  expression  "  20  mi.  an 
hour  "  may  be  understood  as  meaning  that  a  fictitious  automobile 
traveling  in  the  steady  manner  just  described  would  actually 
cover  the  100  mi.  in  just  5  hr. ;  but  for  the  actual  automobile 
which  made  the  trip,  "  20  mi.  an  hour "  gives  only  a  certain 
average  speed. 

So  if  a  man  walks  9  mi.  in  3  hr.,  he  has  an  average  speed  of 
3  mi.  an  hour.  If  a  stone  falls  144  ft.  iii  3  sec,  it  has  an  average 
speed  of  48  ft.  per  second.  In  neither  of  these  cases,  however, 
does  the  average  speed  give  us  any  information  as  to  the  actual 
speed  of  the  moving  object  at  a  given  instant  of  its  motion. 

The  point  we  are  making  is  so  important,  and  it  is  so  often 
overlooked,  that  we  repeat  it  in  the  following  statement: 

If  a  body  traverses  a  distance  in  a  certain  time,  the  average  speed 
of  the  body  in  that  time  is  given  by  the  formula 

,      distance 
average  speed  =  —-. > 

hut  this  formula  does  not  in  general  give  the  true  speed  at  any 
given  time. 


EATES 


EXERCISES 


1.  A  man  runs  a  half  mile  in  2  min.  and  3  sec.  What  is  his 
average  speed  in  feet  per  second  ? 

2.  A  man  walks  a  mile  in  25  min.  What  is  his  average  speed  in 
yards  per  second  ? 

3.  A  train  600  ft.  long  takes  10  sec.  to  pass  a  given  milepost. 
What  is  its  average  speed  in  miles  per  hour  ? 

4.  A  stone  is  thrown  directly  downward  from  the  edge  of  a 
vertical  cliff.  Two  seconds  afterwards  it  passes  a  point  84  ft.  down 
the  side  of  the  cliff,  and  4  sec.  after  it  is  thrown  it  passes  a  point 
296  ft.  down  the  side  of  the  cliff.  What  is  the  average  speed  of  the 
stone  in  falling  between  the  two  mentioned  points  ? 

5.  A  railroad  train  runs  on  the  following  schedule : 


Boston 

10.00  a.m. 

Worcester 

(45  mi.) 

11.10 

Springfield 

(99  mi.) 

12.35  p.m. 

Pittsfield 

(151  mi.) 

2.25 

Albany 

(201  mi.) 

3.55 

Find  the  average  speed  between  each  two  consecutive  stations  and 
for  the  entire  trip. 

6.  A  body  moves  four  times  around  a  circle  of  diameter  6  ft.  in 
1  min.    What  is  its  average  speed  in  feet  per  second  ? 

7.  A  block  slides  from  the  top  to  the  bottom  of  an  inclined 
plane  which  makes  an  angle  of  30°  with  the  horizontal.  If  the  top 
is  50  ft.  higher  than  the  bottom  and  it  requires  §  min.  for  the  block 
to  slide  down,  what  is  its  average  speed  in  feet  per  second  ? 

8.  Two  roads  intersect  at  a  point  C.  B  starts  along  one  road 
toward  C  from  a  point  5  mi.  distant  from  C  and  walks  at  an  average 
speed  of  3  mi.  an  hour.  Twenty  minutes  later  A  starts  along  the 
other  road  toward  C  from  a  point  2  mi.  away  from  C.  At  what 
average  speed  must  A  walk  if  he  is  to  reach  C  at  the  same  instant 
that  B  arrives  ? 

9.  A  man  rows  across  a  river  \  mi.  wide  and  lands  at  a  point 
\  mi.  farther  down  the  river.  If  the  banks  of  the  river  are  parallel 
straight  lines  and  he  takes  ^  hr.  to  cross,  what  is  his  average  speed 
in  feet  per  minute  if  his  course  is  a  straight  line  ? 


SPEED  6 

10.  A  trolley  car  is  running  along  a  straight  street  at  an  average 
speed  of  12  mi.  per  hour.  A  house  is  50  yd.  back  from  the  car  track 
and  100  yd.  up  the  street  from  a  car  station.  A  man  comes  out  of 
the  house  when  a  car  is  200  yd.  away  from  the  station.  What  must 
be  the  average  speed  of  the  man  in  yards  per  minute  if  he  goes  in 
a  straight  line  to  the  station  and  arrives  at  the  same  instant  as 
the  car  ? 

3.  True  speed.  How  then  shall  we  determine  the  speed  at 
which  a  moving  body  passes  any  given  fixed  point  P  in  its 
motion  (Fig.  1)  ?  In  answering  this  question  the  mathema- 
tician begins  exactly  as  does  the  policeman  in  setting  a  trap  for 
speeding.  He  takes  a  point  Q  near  to  P  and  determines  the 
distance  PQ  and  the  time  it  takes  to  pass  over  that  distance. 
Suppose,  for  example,  that  the  distance  P^  is  i  mi.  and  the 
time  is  1  min.  Then,  by  §  2,  the  average  speed  with  which 
the  distance  is  traversed  is 

i  mi.       i  mi.  . 

■f — : —  =  -^ — - —  =  dO  mi.  per  hour. 
1  mm.      Jq  hr. 

This  is  merely  the  average  speed,  however,  and  can  no  more 
be  taken  for  the  true  speed  at  the  point  P  than  could  the  20  mi. 
an  hour  which  we  obtained  by  considering  the  entire  distance 
AB,  It  is  true  that  the  30  mi.  an  hour  obtained  from  the 
interval  P^  is  likely  to  be  nearer  the  true  speed  at  B  than 
was  the  20  mi.  an  hour  obtained  from  AB^  because  the  interval 
P^  is  shorter. 

The  last  statement  suggests  a  method  for  obtaining  a  still 
better  measure  of  the  speed  at  P ;  namely,  by  taking  the  interval 
PQ  still  smaller.  Suppose,  for  example,  that  PQ  i^  taken  as 
y^g  mi.  and  that  the  time  is  6|  sec.  A  calculation  shows  that  the 
average  speed  at  which  this  distance  was  traversed  was  36  mi. 
an  hour.    This  is  a  better  value  for  the  speed  at  P. 

Now,  having  seen  that  we  get  a  better  value  for  the  speed  at 
P  each  time  that  we  decrease  the  size  of  the  interval  PQ,  we 
can  find  no  end  to  the  process  except  by  means  of  the  idea  of  a 
limit  defined  in  §  1.  We  say,  in  fact,  that  the  speed  of  a  moving 
body  at  any  point  of  its  path  is  the  limit  approached  by  the  average 


6  RATES 

speed  computed  for  a  small  distance  heginning  at  that  point,  the 
limit  to  be  determined  by  taking  this  distance  smaller  and  smaller. 

This  definition  may  seem  to  the  student  a  little  intricate,  and 
we  shall  proceed  to  explain  it  further. 

In  the  case  of  the  automobile,  which  we  have  been  using  for 
an  illustration,  there  are  practical  difficulties  in  taking  a  very 
small  distance,  because  neither  the  measurement  of  the  distance 
nor  that  of  the  time  can  be  exact.    This  does  not  alter       q 
the  fact,  however,  that  theoretically  to  determine  the  speed 
of  the  car  we  ought  to  find  the  time  it  takes  to  go  an 
extremely  minute  distance,  and  the  more  minute  the  dis- 
tance the  better  the  result.    For  example,  if  it  were  possi- 
ble to  discover  that  an  automobile  ran  yL-  in.  in  -^gV'o  ^^^•' 
we  should  be  pretty  safe  in  saying  that  it  was  moving  at      \p 
a  speed  of  30  mi.  an  hour.  .  jd 

Such  fineness  of  measurement  is,  of  course,  impossible ; 

but  if  an  algebraic  formula  connecting  the  distance  and 

the  time  is  known,  the  calculation  can  be  made  as  fine  as    ^      „ 

Fig.  2 
this  and  finer.    We  will  therefore  take  a  familiar  case  in 

which  such  a  formula  is  known  ;  namely,  that  of  a  falling  body. 

Let  us  take  the  formula  from  physics  that  if  s  is  the  distance 

through  which  a  body  falls  from  rest,  and  t  is  the  time  it  takes 

to  fall  the  distance  s,  then 

s  =  Ut\  (1) 

and  let  us  ask  what  is  the  speed  of  the  body  at  the  instant 
when  ^  =  2.  In  Fig.  2  let  0  be  the  point  from  which  the  body 
falls,  ij  its  position  when  f  =  2,  and  ij  its  position  a  short  time 
later.  The  average  speed  with  which  the  body  falls  through  the 
distance  P^P^  is,  by  §  2,  that  distance  divided  by  the  time  it 
takes  to  traverse  it.  We  shall  proceed  to  make  several  succes- 
sive calculations  of  this  average  speed,  assuming  ijij  and  the 
corresponding  time  smaller  and  smaller. 

In  so  doing  it  will  be  convenient  to  introduce  a  notation  as 
follows :  Let  t^  represent  the  time  at  which  the  body  reaches  i?, 
and  ^2  the  time  at  which  it  reaches  ij.  Also  let  s^  equal  the 
distance  OP^,  and  s^  the  distance  OP^,    Then  s^—  8^  =  T[Il,  and 


SPEED  7 

t^—  t^  is  the  time  it  takes  to  traverse  the  distance  P^P^,   Then  the 
average  speed  at  which  the  body  traverses  P^P^  is 

(2) 


«.-«! 


Now,  by  the  statement  of  our  particular  problem, 

Therefore,  from  (1),  s^  =  16  (2)^=  64. 

We  shall  assume  a  value  of  t^  a  little  larger  than  2,  compute 
8^  from  (1),  and  the  average  speed  from  (2).  That  having  been 
done,  we  shall  take  t^  a  little  nearer  to  2  than  it  was  at  first,  and 
again  compute  the  average  speed.  This  we  shall  do  repeatedly, 
each  time  taking  t^  nearer  to  2. 

Our  results  can  best  be  exhibited  in  the  form  of  a  table,  as 
follows : 


h 

h 

k  -  h 

h-h 

2.1 
2.01 
2.001 
2.0001 

70.56 
64.6416 
64.064016 
64.00640016 

.1 
.01 
.001 
.0001 

6.56 
.6416 
.064016 
.00640016 

65.6 
64.16 
64.016 
64.0016 

It  is  fairly  evident  from  the  above  arithmetical  work  that  as 
the  time  t^  —  t^  and  the  corresponding  distance  s^  —  s^  become 
smaller,  the  more  nearly  is  the  average  speed  equal  to  64. 
Therefore  we  are  led  to  infer,  in  accordance  with  §  1,  that  the 
speed  at  which  the  body  passes  the  point  ij  is  64  ft.  per  second. 

In  the  same  manner  the  speed  of  the  body  may  be  computed 
at  any  point  of  its  path  by  a  purely  arithmetical  calculation.  In 
the  next  section  we  shall  go  farther  with  the  same  problem  and 
employ  algebra. 

EXERCISES 

1.  Estimate  the  speed  of  a  falling  body  at  the  end  of  the  third 
second,  given  that  5  =  16  ^^  exhibiting  the  work  in  a  table. 

2.  Estimate  the  speed  of  the  body  in  Ex.  1  at  the  end  of  the 
fourth  second,  exhibiting  the  work  in  a  table. 


8  RATES 

3.  The  distance  of  a  falling  body  from  a  fixed  point  at  any  time 
is  given  by  the  equation  s  =  100  -h  16  ^^.  Estimate  the  speed  of  the 
body  at  the  end  of  the  fourth  second,  exhibiting  the  work  in  a  table. 

4.  A  body  is  falling  so  that  the  distance  traversed  in  the  time  t 
is  given  by  the  equation  s  =  16  ^^  +  10 1.  Estimate  the  speed  of  the 
body  when  t  =  2  sec,  exhibiting  the  work  in  a  table. 

5.  A  body  is  thrown  upward  with  such  a  speed  that  at  any  time 
its  distance  from  the  surface  of  the  earth  is  given  by  the  equation 
s  =  100 1  —  16 1^.  Estimate  its  speed  at  the  end  of  a  second,  exhibit- 
ing the  work  in  a  table. 

6.  The  distance  of  a  falling  body  from  a  fixed  point  at  any  time 
is  given  by  the  equation  s  =  50  -f-  20  ^  +  16  ^^  Estimate  its  speed 
at  the  end  of  the  first  second,  exhibiting  the  work  in  a  table. 

4.  Algebraic  method.  In  this  section  we  shall  show  how  it  is 
possible  to  derive  an  algebraic  formula  for  the  speed,  still  con- 
fining ourselves  to  the  special  example  of  the  falling  body  whose 
equation  of  motion  is  s  =  16t^  Cl^ 

Instead  of  taking  a  definite  numerical  value  for  t^^  we  shall 
keep  the  algebraic  symbol  t^.    Then 

s,  =  16t^. 

Also,  instead  of  adding  successive  small  quantities  to  t^  to 
get  t^,  we  shall  represent  the  amount  added  by  the  algebraic 
symbol  h.    That  is,  t  =t  4-h 

and,  from  (1),  8^  =  16  (t^-\-  hy. 

Hence      s.,- s^  =  lQ(t^+ hy-16tl=^2tji+161i\ 

This  is  a  general  expression  for  the  distance  P^I^  in  Fig.  2. 
Now  t^—  t^=  /i,  and  therefore  the  average  speed  with  which  the 
body  traverses  I^I^  is  represented  by  the  expression 

h 

It  is  obvious  that  if  h  is  taken  smaller  and  smaller,  the  aver- 
age speed  approaches  32  ^^  as  a  limit.    In  fact,  the  quantity  32  t^ 


ACCELERATION  9 

satisfies  exactly  the  definition  of  limit  given  in  §  1.  For  if  e 
is  any  number,  no  matter  how  small,  we  have  simply  to  take 
16  A  <  g  in  order  that  the  average  speed  should  differ  from  32  ^^ 
by  less  than  e ;  and  after  that,  for  still  smaller  values  of  A,  this 
difference  remains  less  than  e. 

We  have,  then,  the  result  that  if  the  space  traversed  by  a 
falling  body  is  given  by  the  formula 

the  speed  of  the  body  at  any  time  is  given  by  the  formula 

speed  =  32  ^. 

It  may  be  well  to  emphasize  that  this  is  not  the  result  which 
would  be  obtained  by  dividing  s  by  ^. 

EXERCISE 

Eind  the  speed  in  each  of  the  problems  in  §  3  by  the  method 
explained  in  this  section. 

5.  Acceleration.    Let  us  consider  the  case  of  a  body  which  is 

supposed  to  move  so  that  if  s  is  the  distance  in  feet  and  t  is  the 

time  in  seconds,  ^3  ..,. 

s  =  r.  (1) 

Then,  by  the  method  of  §  4,  we  find  that  if  v  is  the  speed  in 
feet  per  second,  _  q  /2  (<^\ 

We  see  that  when  f  =  1,  v  =  3  ;  when  ^  =  2,  v  =  12  ;  when  t  =  3, 
V  =  27  ;  and  so  on.  That  is,  the  body  is  gaining  speed  with  each 
second.  We  wish  to  find  how  fast  it  is  gaining  speed.  To  find 
this  out,  let  us  take  a  specific  time 

«.=  4. 

The  speed  at  this  time  we  call  v^,  so  that,  by  (2), 

t,^=  3(4)'=  48  ft.  per  second. 

Take  t^—b\ 

then  V  =  3(5y=75  ft.  per  second. 


10 


RATES 


Therefore  the  body  has  gained  75  —  48  =  27  units  of  speed  in 
1  sec.  This  number,  then,  represents  the  average  rate  at  which 
the  body  is  gaining  speed  during  the  particular  second  con- 
sidered. It  does  not  give  exactly  the  rate  at  which  the  speed 
is  increasing  at  the  beginning  of  the  second,  because  the  rate 
is  constantly  changing. 

To  find  how  fast  the  body  is  gaining  speed  when  t^  =  4,  we 
must  proceed  exactly  as  we  did  in  finding  the  speed  itself. 
That  is,  we  must  compute  the  gain  of  speed  in  a  very  small 
interval  of  time  and  compare  that  with  the  time. 

Let  us  take  t^=  4.1. 

Then 

and 


v^=  50.43 
V.  =  2.43. 


Then  the  body  has  gained  2.43  units  of  speed  in  .1  sec,  which 
-^—-  =  24.3  units  per  second. 


is  at  the  rate  of 
Again,  take 
Then 

and 


^^=4.01. 

2;^=  48.2403 
i;  =.2403. 


A  gain  of  .2403  units  of  speed  in  .01  sec.  is  at  the  rate  of 
.2403 


.01 


=  24.03  units  per  second.    We  exhibit  these  results,  and 


one  other  obtained  in  the  same  way,  in  a  table : 


k 

^2 

^2-i^ 

«2  -  ^1 

4.: 

4.01 
4.001 

50.43 

48.2403 

48.024003 

.1 

.01 

.001 

2.43 
.2403 
.024003 

24.3 

24.03 

24.003 

The  rate  at  which  a  body  is  gaining  speed  is  called  its 
acceleration.  Our  discussion  suggests  that  in  the  example  before 
us  the  acceleration  is  24  units  of  speed  per  second.  But  the 
unit  of  speed  is  expressed  in  feet  per  second,  and  so  we  say 
that  the  acceleration  is  24  ft.  per  second  per  second. 


EATE  OF  CHANGE  11 

By  the  method  used  in  determining  speed,  we  may  get  a 
general  formula  to  determine  the  acceleration  from  equation  (2). 
We  take  .      ^    ,   z. 

Then  v^=^(t^+hy 

and  v^  —  v^=  6  t^h+  3  h^. 

The  average  rate  at  which  the  speed  is  gained  is  then 

— i— =  6  ^j  +  3  A, 

and  the   limit  of  this,  as   h  becomes  smaller  and   smaller,   is 
obviously  6  t^. 

This  is,  of  course,  a  result  which  is  valid  only  for  the  special 
example  that  we  are  considering.  A  general  statement  of  the 
meaning  of  acceleration  is  as  follows: 

.       1      ^.  ^^    '^     £  chancre  in  speed 

Acceleration  =  limit  of  — - — ^ — -. — ^, 

change  in  time 

EXERCISES 

1.  If  s  =  4 1^,  find  the  speed  and  the  acceleration  when  t  =^t^. 

2.  If  s  =  ^^  -}-  ^^,  find  the  speed  and  the  acceleration  when  t  =  2. 

3.  If  s  =  3  ^'-^  +  2  ^  +  5,  how  far  has  the  body  moved  at  the  end 
of  the  fifth  second  ?  With  what  speed  does  it  reach  that  point,  and 
how  fast  is  the  speed  increasing  ? 

4.  If  s  =  4  ^^  +  2  ^2  ^  #  +  4,  find  the  distance  traveled  and  the 
speed  when  t  =  2. 

5.  If  s  =  J  f^  +  ^  -|-  10,  find  the  speed  and  the  acceleration  when 
t  =  2  and  when  ^  =  3.  Compare  the  average  speed  and  the  average 
acceleration  during  this  second  with  the  speed  and  the  acceleration 
at  the  beginning  and  the  end  of  the  second. 

6.  li  s  —  at  -^h,  show  that  the  speed  is  constant. 

7.  li  s  =  at^  -\-  ht  -\-  c,  show  that  the  acceleration  is  constant. 

8.  If  s  =  at^ -\- W^ -\- ct -{•  f,  find  the  formulas  for  the  speed  and 
the  acceleration. 

6.  Rate  of  change.  Let  us  consider  another  example  which 
may  be  solved  by  processes  similar  to  those  used  for  determining 
speed  and  acceleration. 


12 


RATES 


A  stone  is  thrown  into  still  water,  forming  ripples  which 
travel  from  the  center  of  disturbance  in  the  form  of  circles 
(Fig.  3).    Let  r  be  the  radius  of  a  circle  and  A  its  area.    Then 

A=TTr\  (1) 

We  wish  to  compare  changes  in  the  area  with  changes  in  the 
radius.  If  we  take  7-^=  3,  then  ^^=  9  tt;  and  if  we  take  r^=  4, 
then  ^2  =  16  TT.  That  is,  a  change 
of  1  unit  in  r,  when  r  =  3,  causes 
a  change  of  7  tt  units  in  A,  We  are 
tempted  to  say  that  A  is  increas- 
ing Ztt  times  as  fast  as  r.  But 
before  making  such  a  statement  it 
is  well  to  see  whether  this  law  holds 
for  all  changes  made  in  /-,  starting 
from  r^  =  3,  and  especially  for  small 
changes  in  r. 

We  will  again  exhibit  the  calcu-  -p^^  o 

lation  in  the  form  of  a  table.    Here 

7-^=3,  ^^=9  7r,  and  r^  is  a  variously  assumed  value  of  r  not 
much  different  from  3. 


^2 

^2 

^2-^1 

A,-A, 

A,-A, 

3.1 

3.01 

3.001 

9.61  TT 

9.0601  TT 
9.006001  IT 

.1 

.01 

.001 

.61  TT 
.0601  TT 

.006001  TT 

6.l7r 

6.01  TT 
6.001  TT 

The  number  in  the  last  column  changes  with  the  number 
r^— r^.  Therefore,  if  we  wish  to  measure  the  rate  at  which  A  is 
changing  as  compared  with  r  at  the  instant  when  r  =  3,  we  must 
take  the  limit  of  the  numbers  in  the  last  column.  That  limit  is 
obviously  6  tt. 

We  say  that  at  the  instant  when  r  =  3,  the  area  of  the  circle  is 
changing  6  ir  times  as  fast  as  the  radius.  Hence,  if  the  radius 
is  changing  at  the  rate  of  2  ft.  per  second,  for  example,  the  area 
is  changing  at  the  rate  of  12  7r  sq.  ft.  per  second.  Another  way 
of  expressing  the  same  idea  is  to  say  that  when  r  =  3,  the  rate 


RATE  OF  CHANGE  13 

of  change  of  A  with  respect  to  r  is  6  tt.  Whichever  form  of  expres- 
sion is  used,  we  mean  that  the  change  in  the  area  divided  by 
the  change  in  the  radius  approaches  a  limit  6  tt. 

The  number  6  tt  was,  of  course,  dependent  upon  the  value 
r  =  3,  with  which  we  started.  Another  value  of  r^  assumed  at 
the  start  would  produce  another  result.  For  example,  we  may 
compute  that  when  r^=  4,  the  rate  of  change  of  A  with  respect 
to  r  is  Stt;  and  when  r^=  5,  the  rate  is  10  tt.  Better  still,  we 
may  derive  a  general  formula  which  will  give  us  the  required 
rate  for  any  value  of  r^. 

To  do  this  take  ,   7 

Then  A^  =  tt  (r^  -\-2r,h-\-  A^) 

and  A^-A^=7r(2r^h-hh^); 

so  that  —^ =  2  7rr,  +  hw. 

'•.-'■1  * 

The  limit  of  this  quantity,  as  h  is  taken  smaller  and  smaller,  is 

2  7rr^, 

Hence  we  see  that  from  formula  (1)  we  may  derive  the  fact 
that  the  rate  of  change  of  A  with  respect  to  r  is  2  irr. 

EXERCISES 

1.  In  the  example  of  the  text,  if  the  circumference  of  the  circle 
which  bounds  the  disturbed  area  is  10  ft.  and  the  circumference  is 
increasing  at  the  rate  of  3  ft.  per  second,  how  fast  is  the  area 
increasing  ? 

2.  In  the  example  of  the  text  find  a  general  expression  for  the 
rate  of  change  of  the  area  with  respect  to  the  circumference. 

3.  A  soap  bubble  is  expanding,  always  remaining  spherical.  If 
the  radius  of  the  bubble  is  increasing  at  the  rate  of  2  in.  per  second;, 
how  fast  is  the  volume  increasing  ? 

4.  In  Ex.  3  find  the  general  expression  for  the  rate  of  change 
of  the  volume  with  respect  to  the  radius. 

5.  If  a  soap  bubble  is  expanding  as  in  Ex.  3,  how  fast  is  the 
area  of  its  surface  increasing? 


14  KATES 

6.  In  Ex.  5  find  the  general  expression  for  the  rate  of  change 
of  the  surface  with  respect  to  the  radius. 

7.  A  cube  of  metal  is  expanding  under  the  influence  of  heat. 
Assuming  that  the  metal  retains  the  form  of  a  cube,  find  the  rate  of 
change  at  which  the  volume  is  increasing  with  respect  to  an  edge. 

8.  The  altitude  of  a  right  circular  cylinder  is  always  equal  to 
the  diameter  of  the  base.  If  the  cylinder  is  assumed  to  expand, 
always  retaining  its  form  and  proportions,  what  is  the  rate  of  change 
of  the  volume  with  respect  to  the  radius  of  the  base  ? 

9.  Find  the  rate  of  change  of  the  area  of  a  sector  of  a  circle  of 
radius  6  ft.  with  respect  to  the  angle  at  the  center  of  the  circle. 

10.  Find  the  rate  of  change  of  the  area  of  a  sector  of  a  circle 
with  respect  to  the  radius  of  the  circle  if  the  angle  at  the  center 

77" 

of  the  circle  is  always  — .    What  is  the  value  of  the  rate  when  the 
radius  is  8  in.  ? 


CHAPTER  II 
DIFFERENTIATION 

7.  The  derivative.  The  examples  we  have  been  considering 
in  the  foregoing  sections  of  the  book  are  alike  in  the  methods 
used  to  solve  them.  We  shall  proceed  now  to  examine  this 
method  so  as  to  bring  out  its  general  character. 

In  the  first  place,  we  notice  that  we  have  to  do  with  two 
quantities  so  related  that  the  value  of  one  depends  upon  the 
value  of  the  other.  Thus  the  distance  traveled  by  a  movmg 
body  depends  upon  the  time,  and  the  area  of  a  circle  depends 
upon  the  radius.  In  such  a  case  one  quantity  is  said  to  be 
a  function  of  the  other.  That  is,  a  quantity  y  is  said  to  be  a 
function  of  another  quantity^  x,  if  the  value  of  y  is  determined  by 
the  value  of  x. 

The  fact  that  y  is  a  function  of  x  is  expressed  by  the  equation 

y  =f(^}^ 
and  the  particular  value  of  the  function  when  x  has  a  definite 
value  a  is  then  expressed  as  /(«).    Thus,  if 

/(2)=  2^- 3(2)^+4(2)4-1  =  5, 
/(0)=0-3(0)  +  4(0)  +  l  =  l. 

It  is  in  general  true  that  a  change  in  x  causes  a  change  in 
the  function  y,  and  that  if  the  change  in  x  is  sufficiently  small, 
the  change  in  y  is  small  also.  Some  exceptions  to  this  may  be 
noticed  later,  but  this  is  the  general  rule.  A  change  in  x  is 
called  an  increment  of  x  and  is  denoted  by  the  symbol  Ax  (read 
"  delta  X ").  Similarly,  a  change  in  y  is  called  an  increment  of 
y  and  is  denoted  by  Ay.    For  example,  consider 

y  =  a^-\-Sx-\-± 

15 


16  DIFFERENTIATION 

When  a;=  2,  y  =  12.  When  a:=  2.1,  ?/ =  12.71.  The  change 
in  X  is  .1,  and  the  change  in  y  is  .71,  and  we  write 

A:r  =  .l,      Ay  =  .71. 

So,  in  general,  if  x^  is  one  value  of  x^  and  x^  a  second  value 

of  x^  then 

Lx  =  x^—  x^^     or     x^=x^-{-Ax;  (1) 

and  if  ?/^  and  y^  are  the  corresponding  values  of  i/,  then 

l^y=y^-yv     ^r     y^=^y^-^Ay,  (2) 

The  word  increment  really  means  "  increase,"  but  as  we  are 
dealing  with  algebraic  quantities,  the  increment  may  be  nega- 
tive when  it  means  a  decrease.  For  example,  if  a  man  invests 
$1000  and  at  the  end  of  a  year  has  $1200,  the  increment  of  his 
wealth  is  $200.  If  he  has  $800  at  the  end  of  the  year,  the 
increment  is  —$200.  So,  if  a  thermometer  registers  65°  in  the 
morning  and  57°  at  night,  the  increment  is  —  8°.  The  incre- 
ment is  always  the  second  value  of  the  quantity  considered  minus 
the  first  value. 

Now,  having  determined  increments  of  x  and  of  ?/,  the  next 
step  is  to  compare  them  by  dividing  the  increment  of  y  by 
the  increment  of  x.  This  is  what  we  did  in  each  of  the  three 
problems  we  have  worked  in  §§  3-6.  In  finding  speed  we  began 
by  dividing  an  increment  of  distance  by  an  increment  of  time, 
in  finding  acceleration  we  began  by  dividing  an  increment  of 
speed  by  an  increment  of  time,  and  in  discussing  the  ripples  in 
the  water  we  began  by  dividing  an  increment  of  area  by  an 
increment  of  radius. 

The  quotient  thus  obtained  is  — ^-    That  is, 

Ay  _  increment  oi  y  _  change  in  y 
Ax      increment  of  x      change  in  x 

An  examination  of  the  tables  of  numerical  values  in  §§  3,  5,  6 

shows  that  the  quotient  — ^  depends  upon  the  magnitude  of  Ax, 

Ax 

and  that  in  each  problem  it  was  necessary  to  determine  its  limit 


DERIVATIVE  17 

as  Arr  approached  zero.    This  limit  is  called  the  derivative  of  y 

with  respect  to  x,  and  is  denoted  by  the  symbol  -^«  We  have  then 

ax 

^  =  limit  of  ^  =  limit  of  2h^5g?4>Ly. 
ax  Ax  change  m  x 

At  present  the  student  is  to  take  the  symbol  -^  not  as  a 

ax 

fraction  but  as  one  undivided  symbol  to  represent  the  deriva- 
tive.   Later  we  shall  consider  what  meaning  may  be  given  to 

dx  and  dy  separately.   At  this  stage  the  form  -^  suggests  simply 

the  fraction  — »  which  has  approached  a  definite  limiting  value. 

A^ 

The  process  of  finding  the  derivative  is  called  differentiation, 
and  we  are  said  to  differentiate  y  with  respect  to  x.  From  the 
definition  and  from  the  examples  with  which  we  began  the  book, 
the  process  is  seen  to  involve  the  following  four  steps : 

1.  The  assumption  at  pleasure  of  Ax. 

2.  The  determination  of  the  corresponding  Ay, 

All 

3.  The  division  of  Ay  by  Ax  to  form  — ^' 

4.  The  determination  of  the  limit  approached  by  the  quotient 
in  step  3  as  the  increment  assumed  in  step  1  approaches  zero. 

Let  us  apply  this  method  to  finding  -^  when  y  =  -'    Let  x^ 

^  ax  X 

be  a  definite  value  of  x,  and  y^=—  the  corresponding  value  of  y, 

1.  Take  Ax  =  h. 
Then,  by  (1),  x^=x^-i-h. 

2.  Then  y=-        ^      " 


x^+h 


whence,  by  (2),     A^  =  ^  -  i  =  -  ^^1^- 

3.  By  division,  — ^  = z — ^ — 

Ax  x^  +  h^x 


18  DIFFERENTIATION 

4.  By  inspection  it  is  evident  that  the  limit,  as  h  approaches 

zero,  is -t  which  is  the  value  of  the  derivative  when  x  =  x.. 

xl 

But  x^  may  be  any  value  oix]  so  we  may  drop  the  subscript  1  and 

write  as  a  general  formula 

^  =  _i. 
dx         x^ 

EXERCISES 

Find  from  the  definition  the  derivatives  of  the  following  ex- 
pressions : 

1.  2/  =  4(0^2  +  1).  5.  y=.x'  +  -^' 

2.  y  =  x'+2x''  +  l.  6.   2/ =  2^' 

3.  y  =  x^  —  x\  7.y  =  ^x^-^^x^  +  x  —  5. 

1 
•^       x^ 

8.  Differentiation  of  a  polynomial.  We  shall  now  obtain  for- 
mulas by  means  of  which  the  derivative  of  a  polynomial  may  be 
written  down  quickly.    In  the  first  place  we  have  the  theorem : 

The  derivative  of  a  polynomial  is  the  sum  of  the  derivatives  of 
its  separate  terms. 

This  follows  from  the  definition  of  a  derivative  if  we  reflect 
that  the  change  in  a  polynomial  is  the  sum  of  the  changes  in  its 
terms.    A  more  formal  proof  will  be  given  later. 

We  have  then  to  consider  the  terms  of  a  polynomial,  which 
have  in  general  the  form  aof.  Since  we  wish  to  have  general 
formulas,  we  shall  omit  the  subscript  1  in  denoting  the  first 
values  of  x  and  y.    We  have  then  the  theorem: 

Ify  =  aaf ,  where  n  is  a  positive  integer  and  a  is  a  constant,  then 

-^  =  an7^-^.  (1) 

dx 

To  prove  this,  apply  the  method  of  §  7 : 

1.   Take  A:c  =  A  ; 

whence  x^—x-\-h. 


POLYNOMIAL  19 

2.  Then  y^=  ax^  =  a(x-\-hy\ 
whence     ^y  =  a(x  +  hy  —  aaf 

3.  By  divison,  ^=^a(nx^-^+^^'^~^^af'-''h+...  +  1f-^\ 

Ax  2  . 

A?/ 

4.  By  inspection,  the  limit  approached  by  —-^  as  A  approaches 
zero,  is  seen  to  be  anx""'^. 

Therefore  -~  =  anx""'^,  as  was  to  be  proved. 
ax 

The  polynomial  may  also  have  a  term  of  the  form  ax.    This 

is  only  a  special  case  of  (1)  with  n  =  1,  but  for  clearness  we 

say  explicitly: 

If  y  =  ax,  where  a  is  a  constant,  then 

^  =  a.  (2) 

dx 

Finally,  a  polynomial  may  have  a  constant  term  c.    For  this 
we  have  the  theorem: 

If  y  =  c,  where  c  is  a  constant,  then 

^  =  0.  (3) 

dx 

The  proof  of  this  is  that  as  c  is  constant,  Ac  is  always  zero, 
no  matter  what  the  value  of  Ax  is.    Hence 

Ax 

dc 
and  therefore  -^-  =  0. 

dx 

As  an  example  of  the  use  of  the  theorems,  consider 
y  =  6a:'H-4a;«-2a:4-7. 


We  write  at  once 


^  =  24a:«+12:r^-2. 
dx 


20  DIFFERENTIATION 

EXERCISES 

Find  the  derivative  of  each  of  the  following  polynomials : 

1.  ic2  +  £c-3.  6.  x'-^7x^-\-21x^-Ux. 

2.  ^£c'-f2ic  +  l.  7.  x}-x^+4:x-l. 

3.  x^-{-^x^+6x^-\-4.x-hl.  8.  3  +  2x2  +  7cc*-lla;^ 

4.  ^x^-{-lx*-\- 2x^+3.  9.  ax^  +  bx''  +  cx-\-f. 

5.  x^-4.x*-^x^-4.x.  10.  a  +  te^ ^  cic^  +  ea;^ 

9.  Sign  of  the  derivative.   If  —  is  positive,  an  increase  in  the 

dx  dy  .  . 

value  of  X  causes  an  increase  in  the  value  of  y.    If  -r-  i^  negative, 

an  increase  in  the  value  of  x  causes  a  decrease  in  the  value  of  y, 

du 
To  prove  this  theorem,  let  us  consider  that  -^  is  positive. 

Then,  since  —  is  the  limit  of  — ^  ?  it  follows  that  —  is  positive 

dx  Ax  Ax 

for  sufficiently  small  values  of  A^:;  that  is,  if  Ax  is  assumed 
positive.  Ay  is  also  positive,   and  therefore    an   increase    of  x 

causes  an  increase  of  y.    Similarly,  if  -^  is  negative,  it  follows 

that  —  is  negative  for  sufficiently  small  values  of  Ax ;  that  is, 

Ax 

if  Ax  is  positive.  Ay  must  be  negative,  so  that  an  increase  of  x 
causes  a  decrease  of  y. 

In  applying  this  theorem  it  is  necessary  to  determine  the 
sign  of  a  derivative.  In  case  the  derivative  is  a  polynomial,  this 
may  be  conveniently  done  by  breaking  it  up  into  factors  and 
considering  the  sign  of  each  factor.  It  is  obvious  that  a  factor 
of  the  form  x—  a  is  positive  when  x  is  greater  than  a,  and 
negative  when  x  is  less  than  a. 

Suppose,  then,  we  wish  to  determine  the  sign  of 
(x  +  S}{x-l}(x-6). 
There  are  three  factors  to  consider,  and  three  numbers  are  im- 
portant ;  namely,  those  which  make  one  of  the  factors  equal  to 
zero.   These  numbers  arranged  in  order  of  size  are  —  3,  1,  and  6. 
We  have  the  four  cases : 

1.  x<—S.  All  factors  are  negative  and  the  product  is 
negative. 


VELOCITY  AND  ACCELERATION  21 

2.  —  3  <  a:  <  1.    The  first  factor  is  positive  and  the  others 
are  negative.    Therefore  the  product  is  positive. 

3.  l<x  <  Q.    The  first  two  factors  are  positive  and  the  last 
is  negative.    Therefore  the  product  is  negative. 

4.  x>  6.    All  factors  are  positive  and  the  product  is  positive. 
As  an  example  of  the  use  of  the  theorem,  suppose  we  have 

and  ask  for  what  values  of  x  an  increase  in  x  will  cause  an 
increase  in  «/.    We  form  the  derivative  and  factor  it.    Thus, 

^  =  3  rr' -  6  a:  -  9  =  3  (a:  + 1 )  (a;  -  3) . 
cix 


Proceeding  as  above,  we  have  the  three  cases 

1.  x<- 
increases  «/. 


1.  X  <  —  l.    -^  is  positive,   and   an   increase  in  x  therefore 
ax 


2.  —  1  <  a:  <  3.  -^  is  negative,  and  therefore  an  increase  in  x 
decreases  y. 

3.  a:  >  3.    -^  is  positive,  and  therefore  an  increase  in  x  in- 

dx 
creases  y. 

These  results  may  be  checked  by  substituting  values  of  x  in 

the  derivative. 

EXERCISES 

Find  for  what  values  of  x  each  of  the  following  expressions  will 
increase  if  x  is  increased,  and  for  what  values  of  x  they  will  decrease 
if  X  is  increased  : 


1. 

a;2_4a;-f5. 

6. 

\x^-\-x^-\hx-\-\\. 

2. 

3a;2+10a;  +  7. 

7. 

x'^-x^-hx^-h. 

3. 

\^-^x-x-. 

8. 

l  +  6x  -f  12a;2+8a;». 

4. 

7_3a^_3a^2 

9. 

6H-6a;  +  6a;2-2x«- 

5. 

2a:3_|_3^2_;12a^ 

+  17. 

10. 

12-12x-6a;2+4a 

10.  Velocity  and  acceleration  (continued).  The  method  by 
which  the  speed  of  a  body  was  determined  in  §  4  was  in  reahty 
a  method  of  differentiation,  and  the  speed  was  the  derivative  of 
the  distance  with  respect  to  the  time.  In  that  discussion,  how- 
ever, we  so  arranged  each  problem  that  the  result  was  positive 


22  DIFFERENTIATION 

and  gave  a  numerical  measure  (feet  per  second,  miles  per  hour, 
etc.)  for  the  rate  at  which  the  body  was  moving.  Since  we  may 
now  expect,  on  occasion,  negative  signs,  we  will  replace  the  word 
speed  by  the  word  velocity^  which  we  denote  by  the  letter  v. 
In  accordance  with  the  previous  work,  we  have 

ds  ,^. 

"  =  71  (^> 

The  distinction  between  speed  and  velocity,  as  we  use  the 
words,  is  simply  one  of  algebraic  sign.  The  speed  is  the  numer- 
ical measure  of  the  velocity  and  is  always  positive,  but  the 
velocity  may  be  either  positive  or  negative. 

From  §  9  the  velocity  is  positive  when  the  body  so  moves  that 
s  increases  with  the  time.  This  happens  when  the  body  moves  in 
the  direction  in  which  s  is  measured.  On  the  other  hand,  the 
velocity  is  negative  when  the  body  so  moves  that  s  decreases 
with  the  time.  This  happens  when  the  body  moves  in  the  direc- 
tion opposite  to  that  in  which  s  is  measured. 

For  example,  suppose  a  body  moves  from  ^  to  ^  (Fig-  1),  a 
distance  of  100  mi.,  and  let  P  be  the  position  of  the  body  at  a 
time  ^,  and  let  us  assume  that  we  know  that  AP  =  4:t.  If  we 
measure  s  from  A^  we  have 

8=AP  =  4:t; 

whence  v  =  —-  =  4:. 

dt 

On  the  other  hand,  if  we  measure  s  from  B^  we  have 

whence  v  =  --  =  —  4. 

dt 

We  will  now  define  acceleration  by  the  formula 

dv 
a  =  —"> 
dt 

in  full  accord  with  §  5 ;  or,  since  v  is  found  by  differentiating  s, 
we  may  write  ^2 


VELOCITY  AND  ACCELERATION  23 

where  the  symbol  on  the  right  indicates  that  8  is  to  be  differ- 
entiated twice  in  succession.  The  result  is  called  a  second 
derivative. 

A  positive  acceleration  means  that  the  velocity  is  increasing, 
but  it  must  be  remembered  that  the  word  increase  is  used  in 
the  algebraic  sense.  For  instance,  if  —  8  is  changed  to  —  5,  it  is 
algebraically  increased,  although  numerically  it  is  smaller.  Hence, 
if  a  negative  velocity  is  increased,  the  speed  is  less.  Similarly, 
if  the  acceleration  is  negative,  the  velocity  is  decreasing,  but  if 
the  velocity  is  negative,  that  means  an  increasing  speed. 

There  are  four  cases  of  combinations  of  signs  which  may 
occur : 

1.  V  positive,  a  positive.  The  body  is  moving  in  the  direction 
in  which  s  is  measured  and  with  increasing  speed. 

2.  V  positive,  a  negative.  The  body  is  moving  in  the  direction 
in  which  s  is  measured  and  with  decreasing  speed. 

3.  V  negative,  a  positive.  The  body  is  moving  in  the  direction 
opposite  to  that  in  which  s  is  measured  and  with  decreasing 
speed. 

4.  V  negative,  a  negative.  The  body  is  moving  in  the  direc- 
tion opposite  to  that  in  which  s  is  measured  and  with  increasing 
speed. 

As  an  example,  suppose  a  body  thrown  vertically  into  the  air 
with  a  velocity  of  96  ft.  per  second.  From  physics,  if  s  is  meas- 
ured up  from  the  earth,  we  have 

8=  96^-16^^ 

From  this  equation  we  compute 

v  =  ^Q-  32  t, 

«  =  -32. 

When  ^  <  3,  V  is  positive  and  a  is  negative.  The  body  is  going 
up  with  decreasing  speed.  When  ^  >  3,  v  is  negative  and  a  is 
negative.    The  body  is  coming  down  with  increasing  speed. 

On  the  other  hand,  suppose  a  body  is  thrown  down  from  a 
height  with  a  velocity  of  96  ft.  per  second.    Then,  if  s  is  measured 


24  DIFFERENTIATION 

down  from  the  point  from  which  the  body  is  thrown,  we  have, 
from  physics,  s  =  96t+16t% 

from  which  we  compute 

v  =  96-\-S'2t, 

a  =  32. 

Here  v  is  always  positive  and  a  is  always  positive.  There- 
fore the  body  is  always  going  down  (until  it  strikes)  with  an 
increasing  speed. 

EXERCISES 

In  the  following  examples  find  the  expression  for  the  velocity 
and  determine  when  the  body  is  moving  in  the  direction  in  which 
s  is  measured  and  when  in  the  opposite  direction  : 

1.  s  =  t^-St-\-e.  3,  s  =  t^-  9^2  -f-  24:t  +  3. 

2.  s  =  10t-t\  4.  s  =  8  +  12t  -  et""  -f-  t\ 

5.  s  =  t'-f-h2. 

In  the  following  examples  find  the  expressions  for  the  velocity 
and  the  acceleration,  and  determine  the  periods  of  time  during  which 
the  velocity  is  increasing  and  those  during  which  it  is  decreasing : 

6.  s  =  3^2  _  4^  ^  4  s.  s  =  ^t^  -  2t^  +  3^  +  2. 

10.  s  =  l-\-At-\-  2^  -  t\ 

11.  Rate  of  change  (continued).  In  §  6  we  have  solved  a 
problem  in  which  we  are  finally  led  to  find  the  rate  of  increase 
of  the  area  of  a  circle  with  respect  to  its  radius.  This  problem 
is  typical  of  a  good  many  others. 

Let  X  be  an  independent  variable  and  y  a  function  of  x, 
A  change  ^x  made  in  x  causes  a  change  A^  in  y.    The  fraction 

~-  compares  the  change  in  y  with  the  change  in  x.  For  exam- 
ple, if  A2:  =  .001,  and  A?/ =.009061,  then  we  may  say  that  the 
change  in  y  is  at  the  average  rate  of  - — — — —  =  9.061  per  unit 

change  in  x.  This  does  not  mean  that  a  unit  change  in  x  would 
actually  make  a  change  of  9.061  units  in  y,  any  more  than  the 


KATE  OF  CHANGE 


25 


statement  that  an  automobile  is  moving  at  the  rate  of  40  mi.  an 
hour  means  that  it  actually  goes  40  mi.  in  an  hour's  time. 

The  fraction  then  gives  a  measure  for  the  average  rate  at 
which  y  is  changing  compared  with  the  change  in  x.  But  this 
measure  depends  upon  the  value  of  Aa:,  as  has  been  shown  in 
the  numerical  calculations  of  §  6.  To  obtain  a  measure  of  the 
instantaneous  rate  of  change  of  y  with  respect  to  x  which  shall 
not  depend  upon  the  magnitude  of  Aa:,  we  must  take  the  limit 

of  — ^,  as  we  did  in  §  6. 

Aa; 

We  have,  therefore,  the  following  definition : 

The  derivative  -j-  measures  the  rate  of  change  of  y  with  respect  to  x. 


Another  way  of  putting  the  same  thing  is  to  say  that  if 


dy 
dx 


has  the  value  m,  then  y  is  changing  m  times  as  fast  as  x. 

Still  another  way  of  expressing  the  same  idea  is  to  say  that 
the  rate  of  change  of  y  with  respect  to  x  is  defined  as  meaning 
the  limit  of  the  ratio  of  a  small  change 
in  y  to  a  small  change  in  x. 

We  will  illustrate  the  above  general 
discussion,  and  at  the  same  time  show 
how  it  may  be  practically  applied,  by  the 
following  example,  which  we  will  first 
solve  arithmetically  and  then  by  calculus. 

Suppose  we  have  a  vessel  in  the  shape  of 
a  cone  (Fig.  4)  of  radius  3  in.  and  altitude 
9  in.  into  which  water  is  being  poured  at 
the  rate  of  100  cu.  in.  per  second.  Re- 
quired the  rate  at  which  the  depth  of  the 
water  is  increasing  when  the  depth  is  6  in. 

From  similar  triangles  in  the  figure,  if  h  is  the  depth  of  the 

water  and  r  the  radius  of  its  surface,  ^  =  o 
of  water. 


Fig.  4 


7r7 


T,\^h'. 


If  V  is  the  volume 
(1) 


We  are  asked  to  find  the  rate  at  which  the  depth  is  increasing 
when  h  is  6  in.    Let  us  call  that  depth  \,  so  that  h^=  6.  Then 


26 


DIFFERENTIATION 


F  =  8  TT.  Now  we  will  increase  h^  by  successive  small  amounts 
and  see  how  great  an  increase  in  V^  is  necessary  to  cause  that 
change  in  h^ ;  that  is,  how  much  water  must  be  poured  in  to  raise 
the  depth  by  that  amount.  The  calculation  may  be  tabulated  as 
follows : 


^h 

AF 

AF 

Ah 

.1 

.01 
.001 

.407  TT 
.04007  TT 
.0040007  TT 

4.07  TT 
4.007  TT 
4.0007  TT 

The  limit  of  the  numbers  in  the  last  column  is  evidently  4  tt. 
Therefore  the  volume  is  increasing  4  tt  times  as  fast  as  the 
depth.  But,  by  hypothesis,  the  volume  is  increasing  at  the  rate 
of  100  cu.  in.  per  second,  so  that  the  depth  is  increasing  at  the 

rate  of  - —  =  7.96  in.  per  second. 
47r  ^ 

We  have  solved  the  problem  by  arithmetic  to  exhibit  again 

the  meaning  of  the  derivative.    The  solution  by  calculus  is  much 

quicker.    We  begin  by  finding 

dV     1     ,. 

This  is  the  general  expression  for  the  rate  of  change  of  V 
with  respect  to  7i,  or,  in  other  words,  it  tells  us  that  V  is  instan- 
taneously increasing  i  irh'^  times  as  fast  as  h  for  any  given  h. 
Therefore,  when  A  =  6,  F  is  increasing  4  tt  times  as  fast  as  A, 
and  as  F  is  increasing  at  the  rate  of  100  cu.  in.  per  second,  h  is 

increasing  at  the  rate  of  - —  =  7.96  in.  per  second. 

EXERCISES 

1.  An  icicle,  which  is  melting,  is  always  in  the  form  of  a  right 
circular  cone  of  which  the  vertical  angle  is  60°.  Find  the  rate  of 
change  of  the  volume  of  the  icicle  with  respect  to  its  length. 

2.  A  series  of  right  sections  is  made  in  a  right  circular  cone  of 
which  the  vertical  angle  is  90°.  How  fast  will  the  areas  of  the  sec- 
tions be  increasing  if  the  cutting  plane  recedes  from  the  vertex  at 
the  rate  of  3  ft.  per  second  ? 


GRAPHS  2T 

3.  A  solution  is  being  poured  into  a  conical  filter  at  the  rate  of 
5  cc.  per  second  and  is  running  out  at  the  rate  of  1  cc.  per  second. 
The  radius  of  the  top  of  the  filter  is  10  cm.  and  the  depth  of  the 
filter  is  30  cm.  Find  the  rate  at  which  the  level  of  the  solution  is 
rising  in  the  filter  when  it  is  one  fourth  of  the  way  to  the  top. 
■  4.  A  peg  in  the  form  of  a  right  circular  cone  of  which  the  ver- 
tical angle  is  60°  is  being  driven  into  the  sand  at  the  rate  of  1  in. 
per  second,  the  axis  of  the  cone  being  perpendicular  to  the  surface 
of  the  sand,  which  is  a  plane.  How  fast  is  the  lateral  surface  of  the 
peg  disappearing  in  the  sand  when  the  vertex  of  the  peg  is  5  in. 
below  the  surface  of  the  sand  ? 

5.  A  trough  is  in  the  form  of  a  right  prism  with  its  ends  equi- 
lateral triangles  placed  vertically.  The  length  of  the  trough  is  10  ft. 
It  contains  water  which  leaks  out  at  the  rate  of  ^  cu.  ft.  per  minute. 
Find  the  rate,  in  inches  per  minute,  at  which  the  level  of  the  water 
is  sinking  in  the  trough  when  the  depth  is  2  ft. 

6.  A  trough  is  10  ft.  long,  and  its  cross  section,  which  is  vertical, 
is  a  regular  trapezoid  with  its  top  side  4  ft.  in  length,  its  bottom 
side  2  ft.,  and  its  altitude  5  ft.  It  contains  water  to  the  depth  of 
3  ft.,  and  water  is  running  in  so  that  the  depth  is  increasing  at  the 
rate  of  2  ft.  per  second.    How  fast  is  the  water  running  in  ? 

7.  A  balloon  is  in  the  form  of  a  right  circular  cone  with  a  hemi- 
spherical top.  The  radius  of  the  largest  cross  section  is  equal  to 
the  altitude  of  the  cone.  The  shape  and  proportions  of  the  balloon 
are  assumed  to  be  unaltered  as  the  balloon  is  inflated.  Find  the 
rate  of  increase  of  the  volume  with  respect  to  the  total  height  of 
the  balloon. 

8.  A  spherical  shell  of  ice  surrounds  a  spherical  iron  ball  concen- 
tric with  it.  The  radius  of  the  iron  ball  is  6  in.  As  the  ice  melts, 
how  fast  is  the  mass  of  the  ice  decreasing  with  respect  to  its  thickness  ? 

12.  Graphs.  The  relation  between  a  variable  x  and  a  function 
y  may  be  pictured  to  the  eye  by  a  graph.  It  is  expected  that 
students  will  have  acquired  some  knowledge  of  the  graph  in 
the  study  of  algebra,  and  the  following  brief  discussion  is  given 
for  a  review. 

Take  two  lines  OX  and  OY  (Fig.  5),  intersecting  at  right 
angles  at  0,  which  is  called  the  origin  of  coordinates.  The  line 
OX  is  called  the  axis  of  x,  and  the  line  OF  the  axis  of  y ;  together 


28 


DIFFERENTIATION 


M 


X 


they  are  called  the  coordinate  axes,  or  axes  of  reference.  On  OX 
we  lay  off  a  distance  OM  equal  to  any  given  value  of  x,  measur- 
ing to  the  right  if  x  is  positive  and  to  the  left  if  x  is  negative. 
From  M  we  erect  a  perpendicular  JfP,  equal  in  length  to  the 
value  of  i/,  measured  up  if  ?/  is  positive  and  down  if  ^  is  negative. 

The  point  P  thus  determined  is  said  to  have  the  coordinates 
X  and  1/  and  is  denoted  by  (x,  y).  It  follows  that  the  numerical 
value   of  X   measures    the    distance  y 

of  the  point  P  from  OF,  and  the 
numerical  value  of  y  measures  the 
distance  of  P  from  OX.  The  coor- 
dinate X  is  called  the  abscissa,  and 
the  coordinate  y  the  ordiiiate.  It  is 
evident  that  any  pair  of  coordinates 
(x,  y)  fix  a  single  point  P,  and  that 
any  point  P  has  a  single  pair  of 
coordinates.  The  point  P  is  said  to 
be  plotted  when  its  position  is  fixed 
in  this  way,  and  the  plotting  is  conveniently  carried  out  on 
paper  ruled  for  that  purpose  into  squares. 

If  «/  is  a  function  of  x,  values  of  x  may  be  assumed  at  pleasure 
and  the  corresponding  values  of  y  computed.  Then  each  pair  of 
values  {x,  «/)  may  be  plotted  and  a  series  of  points  found.  The 
locus  of  these  points  is  a  curve  called  the  graph  of  the  function. 

It  may  happen  that  the  locus  consists  of  distinct  portions  not 

connected  in  the  graph.    In  this  case  it  is  still  customary  to  say 

that  these  portions  together  form  a  single  curve. 

For  example,  let  r  2  ^1^ 

^    '  y=^bx  —  x^.  (1) 

We  assume  values  of  x  and  compute  values  of  y.  The  results 
are  exhibited  in  the  following  table : 


Fig.  5 


X 

-1 

0 

1 

2 

3 

4 

5 

6 

y 

-6 

0 

4 

6 

6 

4 

0 

-6 

These  points  are  plotted  and  connected  by  a  smooth  curve, 
giving  the  result  shown  in  Fig.  6.    This  curve  should  have  the 


GRAPHS 


29 


property  that  the  coordinates  of  any  point  on  it  satisfy  equa- 
tion (1)  and  that  any  point  whose  coordinates  satisfy  (1)  lies 
on  the  curve.  It  is  called  the  graph  both  of  the  function  y  and 
of  the  equation  (1),  and  equation  (1)  is  called  the  equation  of 
the  curve. 

Of  course  we  are  absolutely  sure  of  only  those  points  whose 
coordinates  we  have  actually  computed.  If  greater  accuracy  is 
desired,  more  points  must  be  found 
by  assuming  fractional  values  of  x. 
For  instance,  there  is  doubt  as  to  the 
shape  of  the  curve  between  the  points 
(2,  6)  and  (3,  6).  We  take,  therefore, 
x=2^  and  find  y  =  6^.  This  gives 
us  another  point  to  aid  us  in  draw- 
ing the  graph.  Later,  by  use  of  the 
calculus,  we  can  show  that  this  last 
point  is  really  the  highest  point  of 
the  curve. 

The  curve  (Fig.  6)  gives  us  a 
graphical  representation  of  the  way 
in  which  y  varies  with  x.  We  see,  for 
example,  that  when  x  varies  from  —  1 
to  2,  1/  is  increasing ;  that  when  x 
varies  from  3  to  6,  ?/  is  decreasing;  and  that  at  some  point 
between  (2,  6)  and  (3,  6),  not  yet  exactly  determined,  y  has  its 
largest  value. 

It  is  also  evident  that  the  steepness  of  the  curve  indicates  in 
some  way  the  rate  at  which  i/  is  increasing  with  respect  to  x. 
For  example,  when  x=—l^  an  increase  of  1  unit  in  x  causes  an 
increase  of  6  units  in  y ;  while  when  a;  =1,  an  increase  of  1  unit 
in  X  causes  an  increase  of  only  2  units  in  y.  The  curve  is 
therefore  steeper  when  x=  —1  than  it  is  when  x  =  l. 

Now  we  have  seen  that  the  derivative  ~  measures  the  rate 

ax 

of  change  of  y  with  respect  to  x.   Hence  we  expect  the  derivative 

to  be  connected  in  some  way  with  the  steepness  of  the  curve. 

We  shall  therefore  discuss  this  connection  in  §§14  and  15. 


Fig.  6 


80  DIFFERENTIATION    ' 

EXERCISES 
Plot  the  graphs  of  the  following  equations : 

1.  y  =  2x-}-3.  4.  7j  =  x'^-5x  +  6.         7.y  =  x^ 

2.  ?/  =  -  2a;  +  4.        5.  y  =  x^  +  4:X -{- S.         8.  y  =  x^  -  4x\ 

3.  y  =  5.  6.  y=9-Sx-x\         9.  y  =  x^-l. 

10.  What  is  the  effect  on  the  graph  of  y  =  mx  -\-  S  if  different 
values  are  assigned  to  m  ?  .  How  are  the  graphs  related  ?  What 
does  this  indicate  as  to  the  meaning  of  m  ? 

11.  What  is  the  effect  on  the  graph  oi  y  =  2x  +  b  ii  different 
values  are  assigned  to  ^  ?   What  is  the  meaning  of  b? 

12.  Show  by  similar  triangles  that  y  =  mx  is  always  a  straight 
line  passing  through  0. 

13.  By  the  use  of  Exs.  11  and  12  show  that  y  =  mx  +  bis  always 
a  straight  line. 

13.  Real  roots  of  an  equation.  It  is  evident  that  the  real  roots 
of  the  equation  f(x)  =  0  determine  points  on  the  axis  of  x  at 
which  the  curve  y  =f(x^  crosses  or  touches  that  axis.  More- 
over, if  x^  and  x^  (x^  <  x^)  are  two  values  of  x  such  that  /(a:^) 
and  f(^x^}  are  of  opposite  algebraic  sign,  the  graph  is  on  one  side 
of  the  axis  when  x  =  x^,  and  on  the  other  side  when  x  =  x^. 
Therefore  it  must  have  crossed  the  axis  an  odd  number  of  times 
between  the  points  x  =  x^  and  x  =  x^.  Of  course  it  may  have 
touched  the  axis  at  any  number  of  intermediate  points.  Now,  if 
f(x)  has  a  factor  of  the  form  (x  —  «)*,  the  curve  y  =f(x)  crosses 
the  axis  of  x  at  the  point  x  =  a  when  h  is  odd,  and  touches  the 
axis  of  x  when  k  is  even.  In  each  case  the  equation  f(x)  =  0  is 
said  to  have  k  equal  roots,  x  =  a.  Since,  then,  a  point  of  crossing 
corresponds  to  an  odd  number  of  equal  roots  of  an  equation,  and  a 
point  of  touching  corresponds  to  an  even  number  of  equal  roots, 
it  follows  that  the  equation  f(x)  =  0  has  an  odd  number  of  real 
roots  between  x^  and  x^  if  f(x^  and  /  (x^  have  opposite  signs. 

The  above  gives  a  ready  means  of  locating  the  real  roots  of 
an  equation  in  the  form  f(pc)  =  0,  for  we  have  only  to  find  two 
values  of  x^  as  x^  and  x^^  for  which /(a:)  has  different  signs.  We 
then  know  that  the  equation  has  an  odd  number  of  real  roots 


STRAIGHT  LINE  31 

)etween  these  values,  and  the  nearer  together  x^  and  x^^  the 
more  nearly  do  we  know  the  values  of  the  intermediate  roots. 
In  locating  the  roots  in  this  manner  it  is  not  necessary  to  con- 
struct the  corresponding  graph,  though  it  may  be  helpful. 

Ex.  Find  a  real  root  of  the  equation  x^ •{■  2a:  —  17=  0,  accurate  to  two 
decimal  places. 

Denoting  x^+  2  a;  —  17  hj  f(x)  and  assigning  successive  integral  values 
to  X,  we  find/(2)  =  —  5  and/(3)  =  16.  Hence  there  is  a  real  root  of  the 
equation  between  2  and  3. 

We  now  assign  values  to  x  between  2  and  3,  at  intervals  of  one  tenth, 
as  2.1,  2.2,  2.3,  etc.,  and  we  begin  with  the  values  nearer  2,  since /(2)  is 
nearer  zero  than  is  /(3).  Proceeding  in  this  way  we  find  /(2.3)  =  —  .233 
and/(2.4)  =  1.624 ;  hence  the  root  is  between  2.3  and  2.4. 

Now,  assigning  values  to  x  between  2.3  and  2.4  at  intervals  of  one  hun- 
dredth, we  find  /(2.31)  =  -  .054  and  /(2.32)  =  .127 ;  hence  the  root  is 
between  2.31  and  2.32. 

To  determine  the  last  decimal  place  accurately,  we  let  x  =  2.315  and 
find  /(2.315)  =  .037.  Hence  the  root  is  between  2.31  and  2.315  and  is 
2.31,  accurate  to  two  decimal  places. 

li  /(2.315)  had  been  negative,  we  should  have  known  the  root  to  be 
between  2.315  and  2.32  and  to  be  2.32,  accurate  to  two  decimal  places. 

EXERCISES 

Find  the  real  roots,  accurate  to  two  decimal  places,  of  the  follow- 
ing equations : 

1.  a;8+2a:-6  =  0.  4.  a^*  -  4cc^-f  4  =  0. 

2.  a!^  +  a^-Ml  =  0.  5.  x^  -  Sx^ -{- 6x  -  11  =  0. 

3.  x'-llx-\-5  =  0.  6.  x^+3x^-^4.x-\-7  =  0, 

14.  Slope  of  a  straight  line.    Let  LK  (Figs.  7  and  8)  be  any 

straight  line  not  parallel  to  OX  or  OY,  and  let  I^  (x^j  ^/i)  ^^^ 
ij  (0^2,  y.2)  be  any  two  points  on  it.  If  we  imagine  a  point  to 
move  on  the  line  from  ^  to  ^,  the  increment  of  x  is  x^—x^  and 
the  increment  of  1/  is  y^—  y^-  We  shall  define  the  do'pe  as  the 
ratio  of  the  increment  of  y  to  the  increment  of  x  and  denote  it  by  m. 
We  have  then,  by  definition, 

x^—x.      Ax 


32 


DIFFERENTIATION 


A  geometric  interpretation  of  the  slope  is  easily  given.  For 
if  we  draw  through  ij  a  line  parallel  to  OX,  and  through  i^  a 
line  parallel  to  OY,  and  call  R  the  intersection  of  these  lines, 
then  x^—x^  =  P^R  and  y^—y^  =  RB^.  Also,  if  </>  is  the  angle 
which  the  line  makes  with  OX  measured  as  in  the  figure,  then 


m 


RB, 
P^R 


tan  (^. 


(2) 


It  is  clear  from  the  figures  as  well  as  from  formula  (2)  that 
the  value  of  m  is  independent  of  the  two  points  chosen  to  define 
it,  provided  only  that  these  are  on  the  given  line.  We  may  there- 
fore always  choose  the  two  points  so  that  y^—y^  is  positive. 


: 

T. 

K 

^^ 

y^               1 

■\r- 

^       0 
L 

Ji. 

L 

1 
^2 

it; 

"\< . 

0 

\# 

Fig.  7 


Fig.  8 


Then  if  the  line  runs  up  to  the  right,  as  in  Fig.  7,  x^  —  x^  is 
positive  and  the  slope  is  positive.  If  the  line  runs  down  to  the 
right,  as  in  Fig.  8,  x^  —  x^  is  negative  and  w  is  negative.  There- 
fore the  algebraic  sign  of  m  determines  the  general  direction  in 
which  the  line  runs,  while  the  magnitude  of  m  determines  the 
steepness  of  the  line. 

Formula  (1)  may  be  used  to  obtain  the  equation  of  the  line. 
Let  m  be  given  a  fixed  value  and  the  point  Pi(x-^^  y^  be  held 
fixed,  but  let  i^  be  allowed  to  wander  over  the  line,  taking  on, 
therefore,  variable  coordinates  (x^  ?/).    Equation  (1)  may  then 


be  written 


y-y=m(x-x^. 


(3) 


This  is  the  equation  of  a  line  through  a  fixed  point  (a:^,  y^ 
with  a  fixed  slope  m,  since  it  is  satisfied  by  the  coordinates  of 
any  point  on  the  line  and  by  those  of  no  other  point. 


STRAIGHT  LINE  33 

In  particular,  i^  (x^^,  y^)  may  be  taken  as  the  point  with  coor- 
dinates (0,  6)  in  which  the  line  cuts  0¥,  Then  equation  (3) 
becomes  y  =  mx-it-h  (4) 

Since  any  straight  line  not  parallel  to  OX  ov  to  OY  intersects 
OY  somewhere  and  has  a  definite  slope,  the  equation  of  any 
such  line  may  be  written  in  the  form  (4). 

It  remains  to  examine  lines  parallel  either  to  OX  or  to  OY,  If 
the  line  is  parallel  to  OX,  we  have  no  triangle  as  in  Figs.  7  and  8, 
but  the  numerator  of  the  fraction  in  (1)  is  zero,  and  we  there- 
fore say  such  a  line  has  the  slope  0.    Its  equation  is  of  the  form 

y  =  h,  (5) 

since  it  consists  of  all  points  for  which  this  equation  is  true. 

If  the  line  is  parallel  to  OF,  again  we  have  no  triangle  as  in 
Figs.  7  and  8,  but  the  denominator  of  the  fraction  in  (1)  is  zero, 
and  in  accordance  with  established  usage  we  say  that  the  slope  of 
the  line  is  infinite,  or  that  m  =  oo.  This  means  that  as  the  position 
of  the  line  approaches  parallelism 
with  OY  the  value  of  the  fraction 
(1)  increases  without  limit.  The 
equation  of  such  a  line  is 

x  =  a.  (6) 

Finally  we  notice  that  any  equa- 
tion of  the  form 

Ax  +  By  +  C=0  (7)     '         '        ^'""'^ 

always  represents  a  straight  line.    This  follows  from  the  fact  that 
the  equation  may  be  written  either  as  (4),  (5),  or  (6). 

The  line  (7)  may  be  plotted  by  locating  two  points  and 
drawing  a  straight  line  through  them.  Its  slope  may  be  found 
by  writing  the  equation  in  the  form  (4)  when  possible.  The 
coefficient  of  x  is  then  the  slope. 

If  two  lines  are  parallel  they  make  equal  angles  with  OX. 
Therefore,  if  m^  and  m^  are  the  slopes  of  the  lines,  we  have, 
from  (2),  m=m^,  (8) 

If  two  lines  are  perpendicular  and  make  angles  <^j  and  (f>^ 
respectively  with  OX,  it  is  evident  from  Fig.  9  that  <f>^=  90°-f  <^j ; 


34  DIFFERENTIATION 

whence  tan  (t>^=  —  cot  (j)^= —  •    Hence,  if  m^  and  m^  are  the 

slopes  of  the  lines,  we  have  ^^ 

m,=  --'  (9) 

It  is  easy  to  show,  conversely,  that  if  equation  (8)  is  satis- 
fied by  two  lines,  they  are  parallel,  and  that  if  equation  (9)  is 
satisfied,  they  are  perpendicular.  Therefore  equations  (8)  and 
(9)  are  the  conditions  for  parallelism  and  perpendicularity 
respectively. 

Ex.  1.  Find  the  equation  of  a  straight  line  passing  through  the  point 
(1,  2)  and  parallel  to  the  straight  line  determined  by  the  two  points  (4,  2) 
and  (2,  -  3). 

By  (1)  the  slope  of  the  line  determined  by  the  two  points  (4,  2)  and 

—  3  —  2      5 
(2,  —  3)  is  — —  =  -•    Therefore,  by  (3),  the  equation  of  the  required 

line  is  o       5  /        i  \ 

which  reduces  to  5x  —  2  7/  —  l  =  Q. 

Ex.  2.  Find  the  equation  of  a  straight  line  through  the  point  (2,  —  3) 
and  perpendicular  to  the  line  2a;  —  3?/4-7  =  0. 

The  equation  of  the  given  straight  line  may  be  written  in  the  form 
y  =  §  X  +  ^,  which  is  form  (4).  Therefore  vi  =  §.  Accordingly,  by  (9), 
the  slope  of  the  required  line  is  —  f .  By  (3)  the  equation  of  the  required 
line  is 

y  +  S  =  -%(x-2), 

which  reduces  to  S  x  +  2  y  =  0. 

Ex.  3.  Find  the  equation  of  the  straight  line  passing  through  the  point 
(—  3,  3)  and  the  point  of  intersection  of  the  two  lines  2x  —  y  —  3  =  0  and 
3x-\-2y-l  =  0. 

The  coordinates  of  the  point  of  intersection  of  the  two  given  lines  must 
satisfy  the  equation  of  each  line.  Therefore  the  coordinates  of  the  point 
of  intersection  are  found  by  solving  the  two  equations  simultaneously. 
The  result  is  (1,  -  1). 

We  now  have  the  problem  to  pass  a  straight  line  through  the  points 

3+1 
(—  3,  3)  and  (1,  —  1).  By  (1)  the  slope  of  the  required  line  is  — =  —  1. 

Therefore,  by  (3),  the  equation  of  the  line  is  ""     ~" 

y  +  l  =  -(x-l), 
which  reduces  to  x  +  y  =  0. 


STRAIGHT  LINE  35 


EXERCISES 


1.  Find  the  equation  of  the  straight  line  which  passes  through 
(2,  —  3)  with  the  slope  3. 

2.  Find  the  equation  of  the  straight  line  which  passes  through 
(—  3,  1)  with  the  slope  —  §. 

3.  Find  the  equation  of  the  straight  line  passing  through  the 
points  (1,  4)  and  (f ,  ^). 

4.  Find  the  equation  of  the  straight  line  passing  through  the 
points  (2,  -  3)  and  (-  3,  -  3). 

5.  Find  the  equation  of  the  straight  line  passing  through  the 
point  (2,  —  2)  and  making  an  angle  of  60°  with  OX. 

6.  Find  the  equation  of  the  straight  line  passing  through  the 
point  (^,  —  I)  and  making  an  angle  of  135°  with  OX. 

7.  Find  the  equation  of  the  straight  line  passing  through  the 
point  (—  2,  —  3)  and  parallel  to  the  line  x  -\-2  y  +  1  =  0. 

8.  Find  the  equation  of  the  straight  line  passing  through  the 
point  (—  2,  —  3)  and  perpendicular  to  the  line  3  cc  +  4  z/  —  12  =  0. 

9.  Find  the  equation  of  the  straight  line  passing  through  the 
point  (^,  —  \)  and  parallel  to  the  straight  line  determined  by  the  two 
points  (§,  f )  and  (^,  -  \). 

10.  Find  the  equation  of  the  straight  line  passing  through 
ih  ~  \)  ^^^  perpendicular  to  the  straight  line  determined  by  the 
points  (2,  —  1)  and  (—  3,  5). 

11.  If  )8  is  the  angle  between  two  straight  lines  which  make  angles 
<^j  and  <^2(^2  >  ^i)  respectively  with  OX^  prove  from  a  diagram  similar 
to  Fig.  9  that  /3=  <f>^—  <f>^.  If  tan c^^  =  m^  and  tan  <f>^  =  m^,  prove  by 
trigonometry  that  _ 

tan/8  =  T-f ^• 

12.  Find  the  angle  between  the  lines  x  —  2  ij +  1=0  and 
2a;-32/  +  7  =  0. 

13.  Find  the  angle  between  the  lines  2a;  — 4y  +  5  =  0  and 
5x  +  2y-6  =  0. 

14.  Find  the  angle  between  the  lines  y  =  3ic  +  4  and  x  +  Sy+7  =  0. 

15.  The  vertex  of  a  right  angle  is  at  (2,  —  4)  and  one  of  its  sides 
passes  through  the  point  (—  2,  2).   Find  the  equation  of  the  other  side. 

16.  Find  the  foot  of  the  perpendicular  from  the  origin  to  the  line 
2x- 32/ +  10  =  0. 


36 


DIFFERENTIATION 


15.  Slope  of  a  curve.  Let  AB  (Fig.  10)  be  any  curve  serving 
as  the  graphical  representation  of  a  function  of  x.  Let  i^  be  any 
point  on  the  curve  with  coordinates  x^  =  OM^,  y^=M^P^,  Take 
^=M^M^  and  draw  the  perpendicular  M^^,  fixing  the  point  j? 
on  the  curve  with  the  coor- 
dinates    X.-^  =  OM^,     ^2  =  ^2^' 

Draw    F^E  parallel   to    OX. 
Then 

^E  =  M^M2=Ax, 

Draw    the    straight    line 
^i^,  prolonging  it  to  form  a 


secant  P^S.    Then,  by  §  14, 


^x 


Fig.  10 
is  the  slope  of  the  secant  P^S^  and 

may  be  called  the  average  slope  of  the  curve  between  the  points 
JP  and  i^. 

To  obtain  a  number  which  may  be  used  for  the  actual  slope 
of  the  curve  at  the  point  i^,  it  is  necessary  to  use  the  limit 
process  (with  which  the  student  should  now  be  familiar),  by 
which  we  allow  A^:  to  become  smaller  and  smaller  and  the  point 
ij  to  approach  ij  along  the  curve.  The  result  is  the  derivative 
of  y  with  respect  to  x^  and  we  have  the  following  result: 

The  slope  of  a  curve  at  any  point  is  given  hy  the  value  of  the 
derivative  -j-  at  that  point, 

(XX 

As  this  limit  process  takes  place,  the  point  ij  approaching  the 
point  ij,  it  appears  from  the  figure  that  the  secant  I^S  approaches 
a  limiting  position  P^T.  The  line  iJT  is  called  a  tangent  to  the 
curve,  a  tangent  being  then  by  definition  the  line  approached  as  a 
liinit  by  a  secant  through  two  points  of  the  curve  as  the  two  points 
approach  coincidence.  It  follows  that  the  slope  of  the  tangent  is 
the  limit  of  the  slope  of  the  secant.    Therefore, 

The  slope  of  a  curve  at  any  point  is  the  same  as  the  slope  of  the 
tangent  at  that  point. 


SLOPE  OF  A  CURVE  37 

From  this  and  §  9  we  may  at  once  deduce  the  theorem : 

If  the  derivative  is  positive,  the  curve  runs  up  to  the  right. 
If  the  derivative  is  negative,  the  curve  runs  down  to  the  right.  If 
the  derivative  is  zero,  the  tangent  to  the  curve  is  parallel  to  OX, 
If  the  derivative  is  infinite,  the  tangent  to  the  curve  is  perpendic- 
ular to  OX. 

The  values  of  x  which  make  -^  zero  or  infinite  are  of  par- 

dx 

ticular  interest  in  the  plotting  of  a  curve.  If  the  derivative 
changes  its  sign  at  such  a  point,  the  curve  will  change  its  direc- 
tion from  down  to  up  or  from  up  to  down.  Such  a  point  will 
be  called  a  turning-point.  If  g  is  an  algebraic  polynomial,  its 
derivative  cannot  be  infinite;  so  we  shall  be  concerned  in  this 
chapter  only  with  turning-points  for  which 

^  =  0. 
dx 

They  are  illustrated  in  the  two  following  examples: 

Ex.  1.    Consider  equation  (1)  of  §  12, 

Equating  —  to  zero  and  solving,  we  have  x  =  -  as  a  possible  turning- 
point.    It  is  evident  that  when  x<'-,  ~  is  positive;  and  when  x>-,  -^  is 

^     (Ix  *!i     (IX 

negative.  Therefore  x  =  f  corresponds  to  a  turning-point  of  the  curve  at 
which  the  latter  changes  its  direction  from  up  to  down.  It  may  be  called 
a  high  point  of  the  curve. 

Ex.  2.    Consider 

3,  =  1  (x3  -  3  ^2  -  9  a:  -f-  32). 

Here  g  =  ?(x2  -  2x  -  3)  =  ?(:r  -  3)  (x -f  1). 

Equating  —  to  zero  and  solving,  we  have  x  =  —  1  and  x  =  3  as  possible 

turning-points.    From  the  factored  form  of  -^ ,  and  reasoning  as  in  §  9, 

dx  J 

we  see  that  when  x<  —  l,  —  is  positive  ;  when  —  1<  x  <  3,  -^  is  negative ; 
dx  dx 


Here  ^  =  5-2x  =  2 

dx 


38 


DIFFERENTIATION 


dy 


when  a;  >  3,  -^  is  positive.  Therefore  both  x  =  —  1  and  a:  =  3  give  turning- 
points,  the  former  giving  a  high  point,  and  the  latter  a  low  point. 
Substituting  these  values  of  x  in  the  equa- 
tion of  the  curve,  we  find  the  high  point  to 
be  (—  1,  4|)  and  the  low  point  to  be  (3,  |). 
The  graph  is  shown  in  Fig.  11. 

It  is  to  be  noticed  that  the  solu- 
tions of  the  equation  -^  =  0  do  not 
ax 

always  give   turning-points  as  illus- 
trated in  the  next  example. 


Ex.3.    Consider         y  =  \{x^^ -^  x"^  +  21  x -1%). 


Here 


dy 
dx 


6  a:  -f  9  =  (x  -  3)2. 


dv 
Solving  -^  =  0,  we  have  x  =  3  ;  but  since  the  derivative  is  a  perfect 

square,  it  is  never  negative.   Therefore  x  =  3  does  not  give  a  turning-point, 
although  when  x  =  Z  the  tangent  to  the  curve  is 
parallel  to  OX.    The  curve  is  shown  in  Fig.  12. 

The  equation  of  the  tangent  to  a  curve 
at  a  point  (x^ -,  y^)  is  easily  written  down. 

We  let  \-f]  represent  the  value  of  -j-  at 

.       ^^1  /dy\ 

the  point  (x^^  y^).    Then  m  =  {  —  )',  and, 

from  (3),  §  14,  the  equation  of  the  tangent  is 

Ex.  4.    Find  the  equation  of  the  tangent  at  (1,  —  1)  to  the  curve 
y  =  x^  —  ^  X  ■\-  2. 

We  have 

and 


Fig.  12 


dx 


Therefore  the  equation  of  the  tangent  is 

^4-l=-2(x-l), 
which  reduces  to  2x  -\-  y  —  \  =  0. 


SECOND  DERIVATIVE  39 

From  (2),  §  14,  it  also  follows  that  if  </>  is  the  angle  which 
the  tangent  at  any  point  of  a  curve  makes  with  OX,  then 

J  =  tan<^.  (2) 

EXERCISES 

Locate  the  turning-points,  and  then  plot  the  following  curves : 

2.  2j  =  S-\-Sx-2x\  5.  2/  =  ^(2;:c3  +  3aj2_i2ic-20). 

S.  y  =  x^-Sx'^-\-4:.  6.  y  =  2  -\-9x-\-Sx:'-x^ 

7.  Find  the  equation  of  the  tangent  to  the  curve  y  =  S  —  2x  -\-  x^ 
at  the  point  for  which  x  =  2. 

8.  Find  the  equation  of  the  tangent  to  the  curve  y=l-\-3x—x^—3x^ 
at  the  point  for  which  x  =  —  1. 

9.  Find  the  points  on  the  curve  y  =  x^  -{-  3x^  —  3a;  -f- 1  at  which 
the  tangents  to  the  curve  have  the  slope  6. 

10.  Find  the  equations  of  the  tangents  to  the  curve 

y  =  x^+2x^-x-\-2 
which  make  an  angle  135°  with  OX. 

11.  Find  the  equations  of  the  tangents  to  the  curve  y  =  x^-\-x^—2x 
which  are  perpendicular  to  the  line  Sx-\-2y-\-4:  =  0. 

12.  Find  the  angle  of  intersection  of  the  tangents  to  the  curve 
y  =  x^+x^—2  at  the  points  for  which  x  =  —1  and  x=l  respectively. 

16.  The  second  derivative.    The  derivative  of  the  derivative 
is  called  the  second  derivative  and  is  indicated  by  the  symbol 

^;-(  ,   )  or  -^o'    We  have  met  an  illustration  of  this  in  the 
dx  \dx/  dx 

case  of  the  acceleration.    We  wish  to  see  now  what  the  second 

derivative  means  for  the  graph. 

Since  -j-  is  equal  to  the  slope  of  the  graph,  we  have 


40  DIFFERENTIATION 

From  this  and  §  9  we  have  the  following  theorem : 

If  the  second  derivative  is  positive^  the  slope  is  increasing  as  x 
increases ;  and  if  the  second  derivative  is  negative,  the  slope  is  de- 
creasing as  X  increases. 

We  may  accordingly  use  the  second  derivative  to  distinguish 
between  the  high  turning-points  and  the  low  turning-points  of 
a  curve,  as  follows: 

If,  when  x  =  a,  -^  =  0  and  —^  is  positive,  it  is  evident  that 

d  d 

-^  is  increasing  through  zero ;  hence,  when  x  <  a,  -^  \^  nega- 

dx  ^  dx 

tive,  and  when  x  >  a,  -^  is  positive.    The  point  for  which  x  =  a 
dx 

is  therefore  a  low  turning-point,  by  §  9. 

Similarly,  if,  when  x  =  a,  ^  =  0  and   -— f  is  negative,  it  is 

dy  .  ^^  "^^ 

evident  that  -^  is  decreasing  through  zero ;  hence,  when  x  <  a, 

dy  dy 

~  is  positive,  and  when  x>  a,  -^  is  negative.    The  point  for 

ax  ax 

which  x  =  ais  therefore  a  high  turning-point  of  the  curve,  by  §  9. 
These  conclusions  may  be  stated  as  follows: 

dy  d^y 

If  -^  =  0  and  — ^  is  positive  at  a  point  of  a  curve,  that  point 

.  dy  d\ 

is  a  low  point  of  the  curve.    If  -—  =  0  and  -— |  is  negative  at  a 

ax  ax 

point  of  a  curve,  that  point  is  a  high  point  of  the  curve. 

In  addition  to  the  second  derivative,  we  may  also  have  third, 

fourth,  and  higher  derivatives  indicated  by  the  symbols  -r-f'  -^^ 
etc.    These  have  no  simple  geometric  meaning. 


EXERCISES 

Plot  the  following  curves  after  determining  their  high  and  low 

points  bv  the  use  of  ~  and  ^ : 
ax  dx^ 

1.  2j  =  Sx^-x-2.  3.  y  =  7-lSx-3x^-^4.x\ 

2.  y^S-\-Sx-x'-x\  4.   7/  =  6-15x-\-lSx^-4:X^ 


MAXIMA  AND  MINIMA  41 

17.  Maxima  and  minima.  If  f(a)  is  a  value  of  f(x)  which 
is  greater  than  the  values  obtained  either  by  increasing  or  by 
decreasing  a;  by  a  small  amount,  f(a)  is  called  a  maximum  value 
of  f(x).  If  /(a)  is  a  value  of  f(x)  which  is  smaller  than  the 
values  of  f(x)  found  either  by  increasing  or  by  decreasing  x  by 
a  small  amount, /(a)  is  called  a  minimum  value  oif(x). 

It  is  evident  that  if  we  place 

y  =/(^) 

and  make  the  graph  of  this  equation,  a  maximum  value  oif(x) 
occurs  at  a  high  point  of  the  curve  and  a  minimum  value  at  a 
low  point.  From  the  previous  sections  we  have,  accordingly, 
the  following  rule  for  finding  maxima  and  minima : 

To  find  the  values  of  x  which  give  maximum  or  minimum  values 
of  y^  solve  the  equation 

ax 

If  a:  =  a  is  a  root  of  this  equation,  it  must  be  tested  to  see 
whether  it  gives  a  maximum  or  minimum,  and  which.  We  have 
two  tests: 

Test  I.    If  the  sign  of  -~  changes  from  -^  to  —  as  x  increases 
ax 

through  a,  then  x  =  a  gives  a  maximum  value  of  y.  If  the  sign  of 
-J-  changes  from  —  to  -\-  as  x  increases  through  a,  then  x  —  a  gives 

a  minimum  value  of  y. 

dv  d^v 

Test  II.    If  x  =  a  makes  -^  =  0  and  —~  negative,  then  a:  =  a 

dx  dxr  2 

gives   a  maximum  value  of  y.    If  x  =  a  makes  -^=  0   and  -j^ 

positive,  then  x  =  a  gives  a  minimum  value  of  y. 

Either  of  these  tests  may  be  applied  according  to  convenience. 

It  may  be  noticed  that  Test  I  always  works,  while  Test  II  fails 

d  1/ 
to  give  information  if  — f  =  0  when  x  =  a.    It  is  also  frequently 

possible  by  the  application  of  common  sense  to  a  problem  to 
determine  whether  the  result  is  a  maximum  or  minimum,  and 
neither  of  the  formal  tests  need  then  be  applied. 


42 


DIFFERENTIATION 


Ex.  1.  A  rectangular  box  is  to  be  formed  by  cutting  a  square  from 
each  corner  of  a  rectangular  piece  of  cardboard  and  bending  the  resulting 
figure.  The  dimensions  of  the  piece  of  cardboard  being  20  in.  by  30  in., 
required  the  largest  box  which  can  be  made. 

Let  X  be  the  side  of  the  square  cut  out.  Then,  if  the  cardboard  is  bent 
along  the  dotted  lines  of  Fig.  13,  the  dimensions  of  the  box  are  30  —  2  a:, 
20  —  2  a:,  x.  Let  V  be  the  volume  of  the  box. 
Then         F  =  a:(20  -  2  x)  (30  -  2 x) 

=  600  a;  -  100  x^  +  4  x\ 
dV 


dx 


=  600-200a:  +  12a-2. 
dV 


Equating to  zero,  we  have 

dx 


X 
X 

30 -2  X 

1 

whence 


3  a:2  -  50  ar  +  150  =  0  ; 

25±5V7 


Fig.  13 


3.9  or  12.7. 


The  result  12.7  is  impossible,  since  that  amount  cannot  be  cut  twice 

from  the  side  of  20  in.   The  result  3.9  corresponds  to  a  possible  maximum, 

and  the  tests  are  to  be  applied. 

dV 
To  apply  Test  I  we  write  —  in  the  factored  form 


ilV 
dx 


=  12  (X- 3.9)  (x- 12.7), 


dV 


when  it  appears  that  —  changes  from  +  to  — ,  as  x  increases  through  3.9. 

Hence  t  =  3.9  gives  a  maximum  value  of  V. 

dW 
To  apply  Test  II  we  find  -— 


-200  +  24  a:  and  substitute  x  -  3.9. 


The  result  is  negative.    Therefore  a:  =  3.9  gives  a  maximum  value  of  F. 

The   maximum  value   of    V  is   1056  +  cu.  in.,  found  by  substituting 
a:  =  3.9  in  the  equation  for  F. 


Ex.  2-.  A  piece  of  wood  is  in  the  form  of  a  right  circular  cone,  the 
altitude  and  the  radius  of  the  base  of  which  are  each  equal  to  12  in.  What 
is  the  volume  of  the  largest  right  circular  cylinder  that  can  be  cut  from 
this  piece  of  wood,  the  axis  of  the  cylinder  to  coincide  with  the  axis  of 
the  cone  ? 

Let  X  be  the  radius  of  the  base  of  the  required  cylinder,  y  its  altitude, 
and  V  its  volume.    Then 

\  —  TTX^y.  (1) 

We  cannot,  however,  apply  our  method  directly  to  this  value  of  V,  since 
it  involves  two  variables  x  and  y.   It  is  necessary  to  find  a  connection 


MAXIMA  AND  MINIMA  43 

between  x  and  y  and  eliminate  one  of  them.  To  do  so,  consider  Fig.  14, 
which  is  a  cross  section  of  cone  and  cylinder.   From  similar  triangles  we  have 

FE  ^AD, 
EC~  DC' 
V  12. 


that  is, 


12 -X      12 

whence  y  =  \2  —  x. 

Substituting  in  (1),  we  have 

F  -  12  7rx2  -  Tra:^ 

dV 
whence  —  =  24  ttx  —  3  ttx^. 

dx 

dV 
Equating  —  to  zero  and  solving,  we  find 

dx  B< 

a:  =  0  or  8.  The  value  a:  =  0  is  evidently  not  a 
solution  of  the  problem,  but  x  =  8  is  a  possible 
solution. 

Applying  Test  T,  we  find  that  as  x  increases  through  the  value  8, 

dV 

— —  changes  its  sign  from    +    to   — .     Applying  Test  II,   we  find  that 

dx 

d^V 

— -  =  24:  TT  —  Q  TTx  is  negative  when  x  =  8.    Either  test  shows  that  x  =  8 

dx^ 

corresponds  to  a  maximum  value  of  V.    To  find  V  substitute  a:  =  8  in  the 

expression  for  V.    We  have  V  =  256  rr  cu.  in. 


EXERCISES 

1.  A  piece  of  wire  of  length  20  in.  is  bent  into  a  rectangle  one 
side  of  which  is  x.    Find  the  maximum  area. 

2.  A  gardener  has  a  certain  length  of  wire  fencing  with  which  to 
fence  three  sides  of  a  rectangular  plot  of  land,  the  fourth  side  being 
made  by  a  wall  already  constructed.  Required  the  dimensions  of 
the  plot  which  contains  the  maximum  area. 

3.  A  gardener  is  to  lay  out  a  flower  bed  in  the  form  of  a  sector 
of  a  circle.  If  he  has  20  ft.  of  wire  with  which  to  inclose  it,  what 
radius  will  he  take  for  the  circle  to  have  his  garden  as  large  as 
possible  ? 

4.  In  a  given  isosceles  triangle  of  base  20  and  altitude  10  a  rec- 
tangle of  base  x  is  inscribed.    Find  the  rectangle  of  maximum  area. 

5.  A  right  circular- cylinder  with  altitude  2x  is  inscribed  in  a 
sphere  of  radius  a.    Find  the  cylinder  of  maximum  volume. 


44  DIFFEKENTIATION       • 

6.  A  rectangular  box  with  a  square  base  and  open  at  the  top  is 
to  be  made  out  of  a  given  amount  of  material.  If  no  allowance  is 
made  for  the  thickness  of  the  material  or  for  waste  in  construction, 
what  are  the  dimensions  of  the  largest  box  that  can  be  made  ? 

7.  A  piece  of  wire  12  ft.  in  length  is  cut  into  six  portions,  two 
of  one  length  and  four  of  another.  Each  of  the  two  former  portions 
is  bent  into  the  form  of  a  square,  and  the  corners  of  the  two  squares 
are  fastened  together  by  the  remaining  portions  of  wire,  so  that  the 
completed  figure  is  a  rectangular  parallelepiped.  Find  the  lengths 
into  which  the  wire  must  be  divided  so  as  to  produce  a  figure  of 
maximum  volume. 

8.  The  strength  of  a  rectangular  beam  varies  as  the  product  of 
its  breadth  and  the  square  of  its  depth.  Find  the  dimensions  of  the 
strongest  rectangular  beam  that  can  be  cut  from  a  circular  cylindri- 
cal log  of  radius  a  inches. 

9.  An  isosceles  triangle  of  constant  perimeter  is  revolved  about 
its  base  to  form  a  solid  of  revolution.  What  are  the  altitude  and 
the  base  of  the  triangle  when  the  volume  of  the  solid  generated  is 
a  maximum  ? 

10.  The  combined  length  and  girth  of  a  postal  parcel  is  60  in. 
Find  the  maximum  volume  (1)  when  the  parcel  is  rectangular  with 
square  cross  section ;  (2)  when  it  is  cylindrical. 

11.  A  piece  of  galvanized  iron  Z>  feet  long  and  a  feet  wide  is  to  be 
bent  into  a  U-shaped  water  drain  h  feet  long.  If  we  assume  that  the 
cross  section  of  the  drain  is  exactly  represented  by  a  rectangle  on 
top  of  a  semicircle,  what  must  be  the  dimensions  of  the  rectangle 
and  the  semicircle  in  order  that  the  drain  may  have  the  greatest 
capacity  (1)  when  the  drain  is  closed  on  top  ?  (2)  when  it  is  open 
on  top  ? 

12.  A  circular  filter  paper  10  in.  in  diameter  is  folded  into  a 
right  circular  cone.  Find  the  height  of  the  cone  when  it  has  the 
greatest  volume. 

18.  Integration.  It  is  often  desirable  to  reverse  the  process  of 
differentiation.  For  example,  if  the  velocity  or  the  acceleration 
of  a  moving  body  is  given,  we  may  wish  to  find  the  distance 
traversed ;  or  if  the  slope  of  a  curve  is  given,  we  may  wish  to 
find  the  curve. 


INTEGRATION  45 

The  inverse  operation  to  differentiation  is  called  integration^ 
and  the  result  of  the  operation  is  called  an  integral.  In  the  case 
of  a  polynomial  it  may  be  performed  by  simply  working  the 
formulas  of  differentiation  backwards.    Thus,  if  n  is  a  positive 


integer  and  , 


-f-  =  ax"", 
ax 

then  y-f^l^^'-  (1> 

The  first  term  of  this  formula  is  justified  by  the  fact  that  if  it 
is  differentiated,  the  result  is  exactly  ax"".  The  second  term  is 
justified  by  the  fact  that  the  derivative  of  a  constant  is  zero. 
The  constant  C  may  have  any  value  whatever  and  cannot  be 
determined  by  the  process  of  integration.  It  is  called  the  constant 
of  integration  and  can  only  be  determined  in  a  given  problem  by 
special  information  given  in  the  problem.  The  examples  will 
show  how  this  is  to  be  done. 

Again,  if  ^^^ 

dx 
then  y  =  ax^-C.  (2) 

This  is  only  a  special  case  of  (1)  with  n  =  0. 
Finally,  if 

^  =  ^V^+...+^Va„^  +  ^.  .(3) 

n  -\-l        n  I 

Ex.  1.  The  velocity  v  with  which  a  body  is  moving  along  a  straight 
line  AB  (Fig.  15)  is  given  by  the  equation 


A  B 

How  far  will  the  body  move  in   the  Fig.  15 

time  from  /  =  2  to  /  =  4  ? 

If  when  <  =  2  the  body  is  at  P^,  and  if  when  i  =  4  it  is  at  Pj,  we  are 
to  find  P,P,.  ^^^ 

By  hypothesis,  -^  =  16  ^  +  5. 

Therefore  s  =  8  ^2  +  5  ^  +  C.  (1) 


46 


DIFFERENTIATION 


We  have  first  to  determine  C  If  s  is  measured  from  P^ ,  it  follows  that 
^^^^  t  =  2,        s  =  0. 

Therefore,  substituting  in  (1),  we  have 

0  =  8(2)2  + 5(2) +  C; 
whence  C  =  ~  42, 

and  (1)  becomes  s  =  S  t^  -{■  6  t  -  4:2.  (2) 

This  is  the  distance  of  the  body  from  P^  at  any  time  t.  Accordingly,  it 
remains  for  us  to  substitute  /  =  4  in  (2)  to  find  the  required  distance  PiP^- 
There  results  p^p^  ^  g  ^^^2  +  5  (^4)  _  42  =  106. 

If  the  velocity  is  in  feet  per  second,  the  required  distance  is  in  feet. 


Ex.  2.   Required  the  curve  the  slope  of  which 
at  any  point  is  twice  the  abscissa  of  the  point. 

dy 


By  hypothesis, 


Therefore 


dx 


2x. 


j/  =  x^  +  a 


(1) 


Any  curve  whose  equation  can  be  derived 
from  (1)  by  giving  C  a  definite  value  satisfies 
the  condition  of  the  problem  (Fig.  16).  If  it 
is  required  that  the  curve  should  pass  through 
the  point  (2,  3),  we  have,  from  (1), 

3  =  4  +  C; 
whence  C  =  —  1, 

.and  therefore  the  equation  of  the  curve  is 

y  =  x^  —  1.  Fig.  16 

But  if  it  is  required  that  the  curve  should  pass  through  (—  3,  10), 
we  have,  from  (1),  10  =  9  +  C- 

whence  C  =  1, 

and  the  equation  is  y  =  x^  +  1. 


EXERCISES 

In  the  following  problems  v  is  the  velocity,  in  feet  per  second,  of 
a  moving  body  at  any  time  t. 

1 .  If  V  =  32  ^  +  30,  how  far  will  the  body  move  in  the  time  from 
t  =  2  tot  =  6? 


AREA 


47 


2.  If  V  =  3  ^^  +  4  ^  4-  2,  how  far  will  the  body  move  in  the  time 
from  t  =  ltot  =  3? 

3.  li  V  =  20 1  -\-  25,  how  far  will  the  body  move  in  the  fourth 
second  ? 

4.  It  V  =  t^—  2t  -\-  4:,  how  far  will  the  body  move  in  the  fifth 
and  sixth  seconds  ? 

5.  If  V  =  192  —  32  t,  how  far  will  the  body  move  before  v  =  0? 

6.  A  curve  passes  through  the  point  (1,  —  1),  and  its  slope  at 
any  point  (x,  y)  is  3  more  than  twice  the  abscissa  of  the  point. 
What  is  its  equation? 

7.  The  slope  of  a  curve  at  any  point  (x,  2/)  is  6  £c^  +  2  ic  —  4,  and 
the  curve  passes  through  the  point  (0,  6).   What  is  its  equation  ? 

8.  The  slope  of  a  curve  at  any  point  (cc,  ?/)  is  4  —  3  x  —  a^^,  and 
the  curve  passes  through  the  point  (—  6,  1).    What  is  its  equation  ? 

9.  A  curve  passes  through  the  point  (5,  —  2),  and  its  slope  at  any 
point  (cc,  y)  is  one  half  the  abscissa  of  the  point.  What  is  its  equation  ? 

10.  A  curve  passes  through  the  point  (—2,-4),  and  its  slope  at 
any  point  (a^,  ?/)  is  ic^  —  a?  4- 1.  W^hat  is  its  equation  ? 

19.  Area.    An  important  application  of  integration  occurs  in 
the  problem  of  finding  an  area  bounded  as  follows : 

Let  RS  (Fig.  17)  be  any  curve  with  the  equation  y  —f{x),  and 
let  EB  and  ^C  be  any  two  ordinates.    It  is  required  to  find  the 
area  bounded  by  the  curve  US^        y 
the  two  ordinates  ED  and  BC^ 
and  the  axis  of  x. 

Take  -MP,  any  variable  or- 
dinate between  ED  and  BC^ 
and  let  us  denote  by  A  the 
area  EMPD  bounded  by  the 
curve,  the  axis  of  a;,  the  fixed 
ordinate  ED,  and  the  variable 
ordinate  MP. 

It  is  evident  that  as  values 
are  assigned  to  x=OM,  different  positions  of  MP  and  correspond- 
ing values  of  A  are  determined.    Hence  ^  is  a  function  of  x  for 


which  we  will  find 


dA 
dx 


48  DIFFERENTIATION 

Take  MN=Ax  and  draw  the  corresponding  ordinate  KQ, 
Then  the  area  MNQF=AA.  If  L  is  the  length  of  the  longest 
ordinate  of  the  curve  between  MF  and  NQ,  and  s  is  the  length 
of  the  shortest  ordinate  in  the  same  region,  it  is  evident  that 

sAx  <AA<LAx, 

for  LAx  is  the  area  of  a  rectangle  entirely  surrounding  A  A,  and 
8  Ax  is  the  area  of  a  rectangle  entirely  included  in  A^. 
Dividing  by  Ax,  we  have 

Ax 

As  Ax  approaches  zero,  NQ  approaches  coincidence  with  MP, 
and  hence  s  and  L,  which  are  alwa3^s  between  NQ  and  MP, 
approach  coincidence  with  MP.    Hence  at  the  limit  we  have 

—  =  MP  =  y=f(x).  (1) 

Therefore,  by  integrating, 

A=F(x)  +  C,  (2) 

where  F(x^  is  used  simply  as  a  symbol  for  any  function  whose 
derivative  is  /(a;). 

We  must  now  find  C.  Let  OE  =  a.  When  MP  coincides  with 
ED,  the  area  is  zero.    That  is,  when 

x=  a,  ^  =  0. 

Substituting  in  (2),  we  have 

0=i^(«)  +  C; 

whence  C  =  —  F  (a) , 

and  therefore  (2)  becomes 

A=F(x)-F((i).  (3) 

Finally,  let  us  obtain  the  required  area  EBCD.  If  OB  =  h,  this 
will  be  obtained  by  placing  x  =  h  in  (3).    Therefore  we  have, 

«"^"y'  .  A  =  F(P)-Fia^.  (4) 


AREA 


49 


In  solving  problems  the  student  is  advised  to  begin  with 
formula  (1)  and  follow  the  method  of  the  text,  as  shown  in 
the  following  example: 

Ex.  Find  the  area  bounded  by  the  axis  of  x,  the  curve  y  =  ^x^,  and  the 
ordinates  x  =  1  and  x  =  d. 

In  Fig.  18,  BE  is  the  line  x  =  1,  CD  is  the  line  a:  =  3,  and  the  required 
area  is  the  area  5  CZ>£J.  Then,  by  (1),  / 


dx       3      ' 

. 

L 

D 

/ 

whence         A  =  lx^  +  C. 

/ 

When  x  =  l,  ^  =  0,  and  therefore 

/ 

0  =  l  +  C; 

/ 

f 

whence          C  =  —  ^, 

/ 

and                A  =  lx^-^. 

V 

eX 

Finally,  when 
x  =  Z, 

\ 

Vs 

^^.^ 

0 

B 

C      ' 

^  =  h(^y-h 

=  2|. 

Fig.  18 

EXERCISES 

1.  Find  the  area  bounded  by  the  curve  y  =  4tx  —  x^,  the  axis  of 
x,  and  the  lines  x  =  1  and  £c  =  3. 

2.  Find  the  area  bounded  by  the  curve  y  =  x^ -{- %  x -\- IS,  the 
axis  of  X,  and  the  lines  cc  =  —  6  and  x  =  —  2. 

3.  Find  the  area  bounded  by  the  curve  ?/  =  16  -f  12ic  —  x',  the 
axis  of  Xy  and  the  lines  x  =  —  1  and  x  —  2. 

4.  Find  the  area  bounded  by  the  curve  y  +  cc^  —  9  =  0  and  the 
axis  of  X. 

5.  Find   the   area   bounded   by  the    axis   of   x  and  the   curve 
y  z=  2  X  —  x^. 

6.  Find  the  area  bounded  by  the  curve  y  —  ^x^  —  x^,  the  axis  of 
X,  and  the  line  x  =  1. 

7.  Find  the  area  bounded  by  the  axis  of  x,  the  axis  of  y,  and  the 
curve  4:y  =  x^  —  6^  +  9. 

8.  Find  the  area  bounded  by  the  curve  y  —  x^  —  2 x^  —  A^x  -{•  % 
and  the  axis  of  x. 


50  DIFFERENTIATION 

9.  If  ^  denotes  the  area  bounded  by  the  axis  of  y^  the  curve 
X  =/(^),  a  fixed  line  y  =  h,  and  any  variable  line  parallel  to  OX, 
prove  that 

10.  If  A  denotes  an  area  bounded  above  by  the  curve  y  =f(x), 
below  by  the  curve  y  =  F(x),  at  the  left  by  the  fixed  line  x  =  a, 
and  at  the  right  by  a  variable  ordinate,  prove  that 

20.  Differentials.   The  derivative  has  been  defined  as  the  limit 

of  — ^  and  has  been  denoted  by  the  symbol  -^.    This  symbol 

is  in  the  fractional  form  to  suggest  that  it  is  the  limit  of  a 
fraction,  but  thus  far  we  have  made  no  attempt  to  treat  it  as 
a  fraction. 

It  is,  however,  desirable  in  many  cases  to  treat  the  derivative 
as  a  fraction  and  to  consider  dx  and  dy  as  separate  quantities. 
To  do  this  it  is  necessary  to  define  dx  and  dy  in  such  a  manner 
that  their  quotient  shall  be  the  derivative.  We  shall  begin  by 
defining  dx,  when  x  is  the  independent  variable;  that  is,  the 
variable  whose  values  can  be  assumed  independently  of  any  other 
quantity. 

We  shall  call  dx  the  differential  of  x  and  define  it  as  a  change 
in  X  which  may  have  any  magnitude,  but  which  is  generally 
regarded  as  small  and  may  be  made  to  approach  zero  as  a 
limit.  In  other  words,  the  differential  of  the  independent  variable 
X  is  identical  with  the  increment  of  x;  that  is, 

dx  =  Ax,  (1) 

After  dx  has  been  defined,  it  is  necessary  to  define  dy  so 
that  its  quotient  by  dx  is  the  derivative.    Therefore,  if  y  —f(x) 

and  -^  =f(x),  we  have 

^^  dy=f{x)dx.  (2) 

That  is,  the  differential  of  the  function  y  is  equal  to  the  derivative 
times  the  differential  of  the  independent  variable  x. 


dx 

(1) 

(3) 

X 

aj« 

(2) 

DIFFERENTIALS  51 

In  equation  (2)  the  derivative  appears  as  the  coefficient  of  dx. 
For  this  reason  it  is  sometimes  called  the  differential  coefficient. 

It  is  important  to  notice  the  distinction  between  dy  and  A^. 
The  differential  dy  is  not  the  limit  of  the  increment  A?/,  since  both 
dy  and  A«/  have  the  same  limit,  zero.  Neither  is  dy  equal  to  a  very 
small  increment  Aj/,  since  it  generally  differs  in  value  from  A?/. 
It  is  true,  however,  that  when  dy  and  A?/  both  become  small,  they 
differ  by  a  quantity  which  is  small  compared  with  each  of  them. 
These  statements  may  best  be  under- 
stood from  the  following  examples : 

Ex.  1.   Let  A  be  the  area  of  a  square  with 
the  side  x  so  that 

A  =x\ 

If  a:  is  increased  by  ^x  =  dx,  A  is  increased 
by  A^,  where 

A^  =  (a:  +  dxy  -x'^  =  2xdx+  (dxf.  ' x "dx 

Now,  by  (2),  dA  =  2  xdx,  -pio.  19 

so  that  A^  and  dA  differ  by  (rfa:)^. 

Referring  to  Fig.  19,  we  see  that  dA  is  represented  by  the  rectangles  (1) 
and  (2),  while  A.4  is  represented  by  the  rectangles  (1)  and  (2)  together 
with  the  square  (3) ;  and  it  is  obvious  from  the  figure  that  the  square  (3) 
is  very  small  compared  with  the  rectangles  (1)  and  (2),  provided  dx  is 
taken  small.  For  example,  if  a:  =  5  and  dx  =  .001,  the  rectangles  (1) 
and  (2)  have  together  the  area  2  xdx  =  .01  and  the  square  (3)  has  the 
area  .000001. 

Ex.  2.    Let  s  =  16  f^, 

where  s  is  the  distance  traversed  by  a  moving  body  in  the  time  L 

If  t  is  increased  by  At  =  dt,  we  have  " 

As  =  lQ(t  +  dty  -16fi  =  32tdt-{- 16  (dty, 

and,  from  (2),  ds  =  ^2tdt', 

so  that  As  and  ds  differ  by  16  (dty.  The  term  16  (dt)^  is  very  small  com- 
pared with  the  term  S2tdt,  if  dt  is  small.  For  example,  if  <  =  4  and 
dt  =  .001,  then  S2tdt  =  .128,  while  16  (dty  =  .000016. 

In  this  problem  As  is  the  actual  distance  traversed  in  the  time  dt,  and 
ds  is  the  distance  which  would  have  been  traversed  if  the  body  had  moved 
throughout  the  time  dt  with  the  same  velocity  which  it  had  at  the  begin- 
ning of  the  time  dt. 


52 


DIFFERENTIATION 


In  general,  if  y  =f(pc)  and  we  make  a  graphical  representa- 
tion, we  may  have  two  cases  as  shown  in  Figs.  20  and  21. 

In  each  figure,  M]V  =  PE  =  Ax  =  dx  and  IiQ=Ay,  since  EQ 
is  the  total  change  in  y  caused  by  a  change  of  dx  =  MN  in  x. 
If  FT  is  the  tangent  to  the  curve  at  P,  then,  by  §  15, 


!=/« 


tan  EPT 


so  that,  by  (2),  dy  =  (tan  EFT)  (FE)  =  ET 


Fig.  21 

In  Fig.  20,  dy  <  Ay,  and  in  Fig.  21  dy  >Ay;  but  in  each  case 
the  difference  between  dy  and  Ay  is  represented  in  magnitude 
by  the  length  of  QT. 

This  shows  that  EQ  =Ay  is  the  change  in  y  as  the  point  F  is 
supposed  to  move  along  the  curve  y=f(x),  while  ET=zdy  is 
the  change  in  the  value  of  y  as  the  point  F  is  supposed  to  move 
along  the  tangent  to  that  curve.  Now,  as  a  very  small  arc  does 
not  deviate  much  from  its  tangent,  it  is  not  hard  to  see  graphi- 
cally that  if  the  point  Q  is  tal^en  close  to  P,  the  difference  between 
EQ  and  ET,  namely,  QT,  is  very  small'compared  with  ET. 

A  more  rigorous  examination  of  the  difference  between  the 
increment  and  the  differential  lies  outside  the  range  of  this  book. 

EXERCISES 

1.  If  2/  =  a;«-3cc2  4-4£c  +  l,  find  c?2/. 

2.  li  y  =  x^  ^  ^x^  -  x'^  -\-  Qx,  find  dy. 

3.  If  V  is  the  volume  of  a  cube  of  edge  x,  find  both  AV  and  dV 
and  interpret  geometrically. 


APPROXIMATIONS  53 

4.  If  A  is  the  area  of  a  circle  of  radius  r,  find  both  A.4  and  dA. 
Show  that  A^  is  the  exact  area  of  a  ring  of  width  dr,  and  that  dA 
is  the  product  of  the  inner  circumference  of  the  ring  by  its  width. 

5.  If  F  is  the  volume  of  a  sphere  of  radius  r,  find  AF  and  dV. 
Show  that  A  F  is  the  exact  volume  of  a  spherical  shell  of  thickness 
d)',  and  that  dV  is  the  product  of  the  area  of  the  inner  surface  of 
the  shell  by  its  thickness. 

6.  If  A  is  the  area  described  in  §  19,  show  that  dA  =  ydx.  Show 
geometrically  how  this  differs  from  A^. 

7.  If  s  is  the  distance  traversed  by  a  moving  body,  t  the  time, 
and  V  the  velocity,  show  that  ds  =  vdt.   How  does  ds  differ  from  As  ? 

8.  It  y  =  x^  and  x  =  5,  find  the  numerical  difference  between 
6??/  and  Ay,  with  successive  assumptions  of  dx  =  .01,  dx  =  .001,  and 
dx  =  .0001. 

9.  If.  y  =  x^  and  x  =  3,  find  the  numerical  difference  between  dy 
and  Ay  for  dx  =  .001  and  for  dx  =  .0001. 

10.  For  a  circle  of  radius  4  in.  compute  the  numerical  difference 
between  dA  and  AA  corresponding  to  an  increase  of  r  by  .001  in. 

11.  For  a  sphere  of  radius  3  ft.  find  the  numerical  difference 
between  dV  and  AF  when  r  is  increased  by  1  in. 

21.  Approximations.  The  previous  section  brings  out  the  fact 
that  the  differential  of  y  differs  from  the  increment  of  ^  by  a 
very  small  amount,  which  becomes  less  the  smaller  the  incre- 
ment of  X  is  taken.  The  differential  may  be  used,  therefore,  to 
make  certain  approximate  calculations,  especially  when  the  ques- 
tion is  to  determine  the  effect  upon  a  function  caused  by  small 
changes  in  the  independent  variable.  This  is  illustrated  in  the 
following  examples: 

Ex.  1.  Find  approximately  the  change  in  the  area  of  a  square  of  side 
2  in.  caused  by  a  change  of  .002  in.  in  the  side. 

Let  X  be  the  side  of  the  square,  A  its  area.    Then 
A  =  x^     and     dA  =  2x  dx. 

Placing  x  =  2  and  dx  =  .002,  we  find  dA  =  .008,  which  is  approximately 
the  required  change  in  the  area. 

If  we  wish  to  know  how  nearly  correct  the  approximation  is,  we  may  com- 
pute A^  =  (2.002)2-  (2)2=  .008004,  which  is  the  exact  change  in  A.  Our 
approximate  change  is  therefore  in  error  by  .000004,  a  very  small  amount. 


54  DIFFERENTIATION 

Ex.  2.    Find  approximately  the  volume  of  a  sphere  of  radius  1.9  in. 

The  volume  of  a  sphere  of  radius  2  in.  is  ^^-  ir,  and  the  volume  of  the 
required  sphere  may  be  found  by  computing  the  change  in  the  volume  of 
a  sphere  of  radius  2  caused  by  decreasing  its  radius  by  .1. 

If  r  is  the  radius  of  the  sphere  and  Y  its  volume,  we  have 

V=%irr^     and     dV=^Trr'^dr. 

Placing  r  —  2  and  dr  =  —  .1,  we  find  dV  =  —  1.6  tt.  Hence  the  volume 
of  the  required  sphere  is  approximately 

Aj2  TT- 1.6  7r  =  9.0667  TT. 

To  find  how  much  this  is  in  error  we  may  compute  exactly  the  volume 
of  the  required  sphere  by  the  formula 

'  V=  4  TT  (1.9)3  =9.1453  77. 

The  approximate  volume  is  therefore  in  error  by  .0786  tt,  which  is  less 
than  1  per  cent  of  the  true  volume. 

EXERCISES 

1.  The  side  of  a  square  is  measured  as  3  ft.  long.  If  this  length 
is  in  error  by  1  in.,  find  approximately  the  resulting  error  in  the  area 
of  the  square. 

2.  The  diameter  of  a  spherical  ball  is  measured  as  2i  in.,  and  the 
volume  and  the  surface  are  computed.  If  an  error  of  ^^  in.  has  been 
made  in  measuring  the  diameter,  what  is  the  approximate  error  in 
the  volume  and  the  surface  ? 

3.  The  radius  and  the  altitude  of  a  right  circular  cone  are  meas- 
ured as  3  in.  and  5  in.  respectively.  What  is  the  error  in  the  corre- 
sponding volume  if  an  error  of  -^^  in.  is  made  in  the  radius  ?  What  is 
the  error  in  the  volume  if  an  error  of  ^^^  in.  is  made  in  the  altitude  ? 

4.  Find  approximately  the  volume  of  a  cube  with  3.0002  in.  on 
each  edge. 

5.  The  altitude  of  a  certain  right  circular  cone  is  the  same  as 
the  radius  of  the  base.  Find  approximately  the  volume  of  the  cone 
if  the  altitude  is  3.00002  in. 

6.  The  distance  s  of  a  moving  body  from  a  fixed  point  of  its 
path,  at  any  time  t,  is  given  by  the  equation  s  =  16t'^  -\-  100 1  —  50. 
Find  approximately  the  distance  when  t  =  4.0004. 

7.  Find  the  approximate  value  oi  x^  -\-  x  —  2  when  x  =  1.0003. 

8.  Find  approximately  the  value  of  cc*  +  ^^  +  4  when  x  =  .99989. 


GENERAL  EXERCISES  55 

9.  Show  that  the  volume  of  a  thin  cylindrical  shell  is  approxi- 
mately equal  to  the  area  of  its  inner  surface  times  its  thickness. 

10.  If  F  is  the  volume  and  S  the  curved  area  of  a  right  circular 
cone  with  radius  of  its  base  r  and  its  vertical  angle  2  a,  show  that 
V  =  ^  irr^  ctn  a  and  S  =  in-^  esc  a.  Thence  show  that  the  volume  of 
a  thin  conical  shell  is  approximately  equal  to  the  area  of  its  inner 
surface  multiplied  by  its  thickness. 

GENERAL  EXERCISES 

Find  the  derivatives  of  the  following  functions  from  the  definition 

3  +  2x  1  5.  vi;* 

•     1-x  '  •  x^  +  l'  «  1     * 

6.  — ^• 

a-\-x  rr^  _  1  Va; 

2.  -—^ —  4. 


a-x  '  x'  +  l  7.   Vx^-M.* 

8.  Prove  from  the  definition  that  the  derivative  of  —  is  - — tt- 

9.  By  expanding  and  differentiating,  prove  that  the  derivative 
of  (2ic  +  5)8is  6(2a^  +  5)2. 

10.  By  expanding  and  differentiating,  prove  that  the  derivative 
of  {x^+lfi^  6cc(cc2  +  l)l 

11.  By  expanding  and  differentiating,  prove  that  the  derivative 
of  (x  -j-  a)"  is  n{x  -\-  a)"~^,  where  ti  is  a  positive  integer. 

12.  By  expanding  and  differentiating,  prove  that  the  derivative 
of  (x^  -\-  a^y  is  2  nx(x^  +  a^)"~\  where  ti  is  a  positive  integer. 

13.  Find  when  x'*  +  8  03^  +  24  a?^ -+- 32  a?  +  16  is  increasing  and 
when  decreasing,  as  x  increases. 

14.  Find  when  9  ic^  —  24  a:;*  —  8  a;^  -|-  32  a:  +  11  is  increasing  and 
when  decreasing,  as  x  increases. 

15.  Find  a  general  rule  for  the  values  of  x  for  which  ax^  -{-hx  -{-  c 
is  increasing  or  decreasing,  as  x  increases. 

16.  Find  a  general  rule  for  the  values  of  x  for  which  x^  —  a^x  +  h 
is  increasing  or  decreasing,  as  x  increases. 

17.  A  right  circular  cone  of  altitude  x  is  inscribed  in  a  sphere  of 
radius  a.  Find  when  an  increase  in  the  altitude  of  the  cone  will  cause 
an  increase  in  its  volume  and  when  it  will  cause  a  decrease. 

*  Hint.  In  these  examples  make  use  of  the  relation  VA—Vb=  — =: =• 

■Va  +  Vb 


56  DIFFERENTIATION 

18.  A  particle  is  moving  in  a  straight  line  in  such  a  manner  that 
its  distance  x  from  a  fixed  point  A  of  the  straight  line,  at  any  time  t, 
is  given  by  the  equation  x  =  t^  —  ^t"^  -\-l^t  -{- 100.  When  will  the 
particle  be  approaching  A  ? 

19.  The  velocity  of  a  certain  moving  body  is  given  by  the  equation 
V  =  t"^  —  7  t  -\-10.  During  what  time  will  it  be  moving  in  a  direction 
opposite  to  that  in  which  s  is  measured,  and  how  far  will  it  move  ? 

20.  If  a  stone  is  thrown  up  from  the  surface  of  the  earth  with  a 
velocity  of  200  ft.  per  second,  the  distance  traversed  in  t  seconds  is 
given  by  the  equation  s  =  200  ^  —  16  ^^.  Find  when  the  stone  moves 
up  and  when  down. 

21.  The  velocity  of  a  certain  moving  body,  at  any  time  t,  is  given 
by  the  equation  v  =  t"^  —  ^t  -\-12.  Find  when  the  velocity  of  the 
body  is  increasing  and  when  decreasing. 

22.  At  any  time  t,  the  distance  s  of  a  certain  moving  body  from  a 
fixed  point  in  its  path  is  given  by  the  equation  s  =  16  —  24  ^  -|-  9  if ^  —  ^^ 
When  is  its  velocity  increasing  and  when  decreasing  ?  When  is  its 
speed  increasing  and  when  decreasing  ? 

23.  At  any  time  #,  the  distance  of  a  certain  moving  body  from  a 
fixed  point  in  its  path  is  given  by  the  equation  s  =  ^*  —  6  ^^  -f  9  ^  +  1. 
When  is  its  speed  increasing  and  when  decreasing  ? 

24.  A  sphere  of  ice  is  melting  at  such  a  rate  that  its  volume  is 
decreasing  at  the  rate  of  10  cu.  in.  per  minute.  At  what  rate  is  the 
radius  of  the  sphere  decreasing  when  the  sphere  is  2  ft.  in  diameter  ? 

25.  Water  is  running  at  the  rate  of  1  cu.  ft.  per  second  into  a 
basin  in  the  form  of  a  frustum  of  a  right  circular  cone,  the  radii  of 
the  upper  and  the  lower  base  being  10  in,  and  6  in.  respectively,  and 
the  depth  being  6  in.  How  fast  is  the  water  rising  in  the  basin  when 
it  is  at  the  depth  of  3  in.  ? 

26.  A  vessel  is  in  the  form  of  an  inverted  right  circular  cone  the 
vertical  angle  of  which  is  60°.  The  vessel  is  originally  filled  with 
liquid  which  flows  out  at  the  bottom  at  the  rate  of  3  cu.  in.  per  minute. 
At  what  rate  is  the  inner  surface  of  the  vessel  being  exposed  when 
the  liquid  is  at  a  depth  of  1  ft.  in  the  vessel  ? 

27.  Find  the  equation  of  the  straight  line  which  passes  through 
the  point  (4,  —  5)  with  the  slope  —  \. 

28.  Find  the  equation  of  the  straight  line  through  the  points 
(2,-3)  and  (-3,  4). 


GENERAL  EXERCISES  57 

29.  Find  the  equation  of  the  straight  line  determined  by  the  points 

(2,  -  4)  and  (2,  4). 

30.  Find  the  equation  of  the  straight  line  through  the  point 
(1,  —  3)  and  parallel  to  the  line  ic  —  2?/-|-7  =  0. 

31.  Find  the  equation  of  the  straight  line  through  the  point 
(2,  7)  and  perpendicular  to  the  line  2x  -\-  4:y  -{-  9  =  0. 

32.  Find  the  angle  between  the  straight  lines  2x  -\-  Sy  -[-  5  =  0 
and  y-\-Sx-{-l  =  0. 

Find  the  turning-points  of  the  following  curves  and  draw  the 
graphs : 

33.  y  =  S-x-x\ 

34.  y  =  16x^-4:0x-^25, 
S5.   ij  =  ^(x^-  4.x-2y 

36.  y  =  x^-6x^-15x-h5. 

37.  y  =  x^-\-Sx^-9x-U. 

38.  Find  the  point  of  intersection  of  the  tangents  to  the  curve 
y  =  x  -\-  2x^  —  x^  at  the  points  for  which  x  =  —  1  and  x  =  2 
respectively. 

39.  Show  that  the  equation  of  the  tangent  to  the  curve  y  =  ax"^ 
-\-  2bx  -^  c  Sit  the  point  (xi,  yi)  is  y  =  2  (ax^ -\-h)x  —  ax^ -\-  c. 

40.  Show  that  the  equation  of  the  tangent  to  the  curve  y  =  x^ 
-\-  ax  -\-  b  at  the  point  (x^ ,  y^)  is  y  =  (Sx^  -\-  a)x  —  2 xf  +  b. 

41.  Find  the  area  of  the  triangle  included  between  the  coordinate 
axes  and  the  tangent  to  the  curve  y  =  x^  Sit  the  point  (2,  8). 

42.  Find  the  angle  between  the  tangents  to  the  curve  y  =  2x^ 
■i-  4:x^  —  X  at  the  points  the  abscissas  of  which  are  —  1  and  1 
respectively. 

43.  Find  the  equation  of  the  tangent  to  the  curve  y  =  x^^Sx'^ 
4-  4  ic  —  12  which  has  the  slope  1. 

44.  Find  the  points  on  the  curve  y  =  Sx'^  —  Ax^  at  which  the 
tangents  are  parallel  to  the  line  x  —  y  =  0. 

45.  A  length  I  of  wire  is  to  be  cut  into  two  portions  which  are  to 
be  bent  into  the  forms  of  a  circle  and  a  square  respectively.  Show 
that  the  sum  of  the  areas  of  these  figures  will  be  least  when  the  wire 
is  cut  in  the  ratio  tt  :  4. 


58  DIFFERENTIATION 

46.  A  log  in  the  form  of  a  frustum  of  a  cone  is  10  ft.  long,  the 
diameters  of  the  bases  being  4  ft.  and  2  ft.  A  beam  with  a  square 
cross  section  is  cut  from  it  so  that  the  axis  of  the  beam  coincides 
with  the  axis  of  the  log.  Find  the  beam  of  greatest  volume  that 
can  be  so  cut. 

47.  Required  the  right  circular  cone  of  greatest  volume  which  can 
be  inscribed  in  a  given  sphere. 

48.  The  total  surface  of  a  regular  triangular  prism  is  to  be  k.  Find 
its  altitude  and  the  side  of  its  base  when  its  volume  is  a  maximum. 

49.  A  piece  of  wire  9  in.  long  is  cut  into  five  pieces,  two  of  one 
length  and  three  of  another.  Each  of  the  two  equal  pieces  is  bent 
into  an  equilateral  triangle,  and  the  vertices  of  the  two  triangles  are 
connected  by  the  remaining  three  pieces  so  as  to  form  a  regular 
triangular  prism.  How  is  the  wire  cut  when  the  prism  has  the  largest 
volume  ? 

50.  If  t  is  time  in  seconds,  v  the  velocity  of  a  moving  body  in 
feet  per  second,  and  v  —  200  —  32  t,  how  far  will  the  body  move  in 
the  first  5  sec.  ? 

51.  li  V  =  200  —  32  t,  where  v  is  the  velocity  of  a  moving  body 
in  feet  per  second  and  t  is  time  in  seconds,  how  far  will  the  body 
move  in  the  fifth  second  ? 

52.  A  curve  passes  through  the  point  (2,  —  3),  and  its  slope  at 
any  point  is  equal  to  3  more  than  twice  the  abscissa  of  the  point. 
Find  the  equation  of  the  curve. 

53.  A  curve  passes  through  the  point  (0,  0),  and  its  slope  at  any 
point  is  ic^  —  2  a?  -h  7.    Find  its  equation. 

54.  Find  the  area  bounded  by  the  curve  ?/  +  a?^  —  16  =  0  and  the 
axis  of  X. 

55.  Find  the  area  bounded  by  the  curve  y  =  2  x^  —  1^  x^ -\-  ^Q>  x  +1 
and  the  ordinates  through  the  turning-points  of  the  curve. 

56.  Find  the  area  between  the  curve  y  —  x?  and  the  straight  line 
2/  =  X  -f  6. 

57.  Find  the  area  between  the  curves  y  =  x'^  and  y  —  1%  —  x^. 

58.  The  curve  y  =  ax?  is  known  to  jiass  through  the  point  (A,  Iz). 
Prove  that  the  area  bounded  by  the  curve,  the  axis  of  x,  and  the 
line  x  =  li\^  \  hk. 

59.  Compute  the  difference  between  A^  and  dA  for  the  area  A  of 
a  circle  of  radius  5,  corresponding  to  an  increase  of  .01  in  the  radius. 


GENERAL  EXERCISES  59 

60.  Compute  the  difference  between  A  F  and  dV  for  the  volume  V 
of  a  sphere  of  radius  5,  corresponding  to  an  increase  of  .01  in  the 
radius. 

61.  If  a  cubical  shell  is  formed  by  increasing  each  edge  of  a  cube 
by  dx,  show  that  the  volume  of  the  shell  is  approximately  equal  to 
its  inside  surface  multiplied  by  its  thickness. 

62.  If  the  diameter  of  a  sphere  is  measured  and  found  to  be  2  ft., 
and  the  volume  is  calculated,  what  is  the  approximate  error  in  the 
calculated  volume  if  an  error  of  i  in.  has  been  made  in  obtaining 
the  radius  ? 

63.  A  box  in  the  form  of  a  right  circular  cylinder  is  6  in.  deep 
and  6  in.  across  the  bottom.  Find  the  approximate  capacity  of  the 
box  when  it  is  lined  so  as  to  be  5.9  in.  deep  and  5.9  in.  across  the 
bottom. 

64.  A  rough  wooden  model  is  in  the  form  of  a  regular  quadran- 
gular pyramid  3  in.  tall  and  3  in.  on  each  side  of  the  base.  After  it 
is  smoothed  down,  its  dimensions  are  all  decreased  by  .01.  What  is 
the  approximate  volume  of  the  material  removed  ? 

65.  By  use  of  the  differential  find  approximately  the  area  of  a 
circle  of  radius  1.99.   What  is  the  error  made  in  this  approximation  ? 

66.  Find  approximately  the  value  of  x^+  4  ic^-f-  x  when  x  =  3.0002 
and  when  x  =  2.9998. 

67.  The  edge  of  a  cube  is  2.0001  in.  Find  approximately  its  surface. 

68.  The  motion  of  a  certain  body  is  defined  by  the  equation 
s  =  f  -\-  3f  -\-  9t  —  27.  Find  approximatel}^  the  distance  traversed 
in  the  interval  of  time  from  ^  =  3  to  t  =  3.0087. 


CHAPTER  III 
SUMMATION 

22.  Area  by  summation.  Let  us  consider  the  problem  to  find 
the  area  bounded  by  the  curve  y  =  ^x^,  the  axis  of  x^  and  the 
ordinates  x=2  and  a;  =  3  (Fig.  22).  This  may  be  solved  by  the 
method  of  §  19 ;  but  we  wish  to  show  that  it  may  also  be  considered 
as  a  problem  in  summation,  since  the  area  is  approximately  equal 
to  the  sum  of  a  number  of  rectangles  constructed  as  follows : 

We  divide  the  axis  of  x  between  x  =  2  and  2^  =  3  into  10  parts, 

3  —  2 

each  of  which  we  call  Aa:,  so  that  A2:  =  --— —  =.1.    If  x^  is  the 

first  point  of  division,  x^  the  second  point,  and  so  on,  and 
rectangles  are  constructed  as  shown  in  the  figure,  then  the 
altitude  of  the  first  rectangle  is  ^(2)^  that  of  the  second  rec- 
tangle is  ^  xl  =  \  (2.iy  =  .^^%  and  so  on.  The  area  of  the 
first  rectangle  is  i(2)^Aa7  =.08,  that  of  the  second  rectangle  is 


^j^Aa;  =  1(2-1)"^  =  -0882, 

and  so  on. 

Accordingly,  we  make  the 

i  following  ca 

Iculatii 

^  =  2, 

\(.^y^-= 

.08 

x=2.\. 

•(^,)^Ax  = 

.0882 

x  =  2.2, 

w.r^= 

.0968 

0^3=  2.3, 

i(x^yAx= 

.1058 

x=  2.4, 

K^,yA^  = 

.1152 

x^=  2.5, 

i(^5)'^  = 

.1250 

^e=2-6, 

K^e)'^^  = 

.1352 

^,=  2.7, 

l(x^fAx  = 

.1458 

^,=  2.8, 

U^fAx^ 

.1568 

x,=  2.9, 

lCx^yAx  = 

.1682 

1.2170 

This  is  a  first  approximation  to  the  area. 

For  a  better  approximation  the  axis  of  x  between  x=  2  and  x=  3 
may  be  divided  into  20  parts  with  A.3:;=.05.   The  result  is  1.2418. 

60 


AREA 


61 


If  the  base  of  the  required  figure  is  divided  into  100  parts  with 
Aa:=.01,  the  sum  of  the  areas  of  the  100  rectangles  constructed 
as  above  is   1.26167. 

The  larger  the  num- 
ber of  parts  into  which 
the  base  of  the  figure  is 
divided,  the  more  nearly 
is  the  required  area  ob- 
tained. In  fact,  the  re- 
quired area  is  the  limit 
approached  as  the  number 
of  parts  is  indefinitely 
increased  and  the  size  of 
Ax  approaches  zero. 

We  shall  now  proceed  to  generalize  the  problem  just  handled. 
Let  LK  (Fig.  23)  be  a  curve  with  equation  i/  =f(x)^  and  let 
OE=a  and  OB  =  h.    It  is  re- 
quired to  find  the  area  bounded 
by  the  curve  LK^  the  axis  of  x^ 
and  the  ordinates  at  E  and  B. 

For  convenience  we  assume 
in  the  first  place  that  a  <h 
and  that  f(x)  is  positive  for 
all  values  of  x  between  a  and  h. 
We  will  divide  the  line  EB 
into  n  equal  parts  by  placing 


Fig. 22 


Ax  = 


and  laying  off  the 


M,  3/2  3/s  3/4  3/,  il/e  3/,  3/8 
Fig.  23 


'=M^_^B=Ax  (in  Fig.  23,  w  =  9). 
C>^X,-i=^„-i-  Draw  JS-i)  =/(«), 
^/„_i^_i=/(^„-i).  and^C;  also 


lengths  EM^ =M^M^= M^M^  = 
Let  0M^=  x^,  0M^=  x.^,  • 

M,P,=f(x,^.M,P,=f(x,),  . 

DR^,  P^R^,  P^R,,  .  .  .,  Pn-^Rn^  parallel  to  OX.    Then 
/(a)Aa;  =  the  area  of  the  rectangle  EDR^M^, 
f(x;)Ax  =  the  area  of  the  rectangle  M^P^R^M.;,, 
f(x^Ax  =  t\\Q  area  of  the  rectangle  M^P^R^M^, 

/(:r„_i)Aa:  =  the  area  of  the  rectangle  M„_,P„_,R,B. 


62  SUMMATION 

The  sum 

/(«)A:r  +f(x,)^x  +f(x.^^x  +  .  .  .  +f(x,^^,)^x  (1) 

is  then  the  sum  of  the  areas  of  these  rectangles  and  equal  to 
the  area  of  the  polygon  EDE^I^R.^  .  •  .  Ii^^_^]^^_^R^B,  It  is  evident 
that  the  limit  of  this  sum  as  n  is  indefinitely  increased  is  the 
area  bounded  by  ED^  EB,  BC,  and  the  arc  DC. 

The  sum  (1)  is  expressed  concisely  by  the  notation 

1=0 

where  2  (sigma),  the  Greek  form  of  the  letter  S,  stands  for  the 
word  ''  sum,"  and  the  whole  expression  indicates  that  the  sum 
is  to  be  taken  of  all  terms  obtained  from  /(a:.) Aa:  by  giving  to  i 
in  succession  the  values  0,  1,  2,  3,    •  •  .,  n  — 1,  where  x^=  a. 
The  limit  of  this  sum  is  expressed  by  the  symbol 


/(x^dx, 

Ja 

where  /  is  a  modified  form  of  S. 

f(x)dx  =  Lim  V/(a:.)A:r  =  the  area  EBCD. 

a  «=»    (=0 

It  is  evident  that  the  result  is  not  vitiated  if  ED  or  BC  is  of 
length  zero. 

23.  The  definite  integral.  We  have  seen  in  §  19  that  if  A  is 
the  area  EBCD  of  §  22, 

A  =  E(b}-E(a), 

where  E(^x^  is  any  function  whose  derivative  is /(a:).   Comparing 
this  with  the  result  of  §  22,  we  have  the  important  formula 

f''f(xyx=FCb)-FC<i-).  (1) 

\J  a 

The  limit  of  the  sum  (1),  §  22,  which  is  denoted  by  /  f(x)dx, 
is  called  a  definite  integral,  and  the  numbers  a  and  h  are  called 


DEFINITE  INTEGRAL  63 

the  lower  limit  and  the  upper  limit*,  respectively,  of  the  definite 
integral. 

On  the  other  hand,  the  symbol  I  f(x)  dx  is  called  an  indefinite 

integral  and  indicates  the  process  of  integration  as  already  de- 
fined in  §  18. 

Thus,  from  that  section,  we  have 


/ 


ax^'dx  = +  C, 

71  +  1 


/ 


adx  =  ax  +  C^ 


and,  in  general,         \  f  (x)  dx  =  F (x)  +  C, 

where  F(x)  is  any  function  whose  derivative  is /(a:). 

We  may  therefore  express  formula  (1)  in  the  following  rule : 

To  find  the  value  of  I  f(x)dx,  evaluate    lf(^x^dx,  substitute 

x  =  h  and  x  —  a  successively,  and  subtract  the  latter  result  from 
the  former. 

It  is  to  be  noticed  that  in  evaluating  \  f(x)dx  the  constant 

of  integration  is  to  be  omitted,  because  if  it  is  added,  it  dis- 
appears in  the  subtraction,  since 

iF(b)  +  C]  -  [i^(«)  +  C]  =  F(b)  -  F{ay 

In  practice  it  is  convenient  to  express  F(J))  —  F(^a')  by  the 
symbol  [i^(^)]«,  so  that 


X 


f(x)dx  =  [F(x)}^. 


Ex.  1.  The  example  of  §  22  may  now  be  completely  solved.  The  required 

area  is 

/,3^2  r^3-|3     97      s      in 


J2    5'^"  Ll5j2~15      15~15 


*  The  student  should  notice  that  the  word  'Mimit"  is  here  used  in  a  sense 
quite  different  from  that  in  which  it  is  used  when  a  variable  is  said  to  approach 
a  limit  (§1). 


64 


SUMMATION 


The  expression /(:r)  c?a;  which  appears  in  formula  (1)  is  called 
the  element  of  integration.  It  is  obviously  equal  to  dF(x),  In 
fact,  it  follows  at  once  from  §  19  that 

dA  =  ydx  =f(jc)  dx. 

In  the  discussion  of  §  22  we  have  assumed  that  y  and  dx  are 
positive,  so  that  dA  is  positive.  If  y  is  negative  —  that  is,  if  the 
curve  in  Fig.  23  is  below  the  axis  oi  x  —  and  if  dx  is  positive, 
the  product  ydx  is  negative  and  the 
area  found  by  formula  (1)  has  a  nega- 
tive sign.  Finally,  if  the  area  required 
is  partly  above  the  axis  of  x  and 
partly  below,  it  is  necessary  to  find 
each  part  separately,  as  in  the  follow- 
ing example: 

Ex.  2.  Find  the  area  bounded  by  the 
curve  y  ■=  x^  —  x^  —  Q  X  and  the  axis  of  x. 

Plotting  the  curve  (Fig.  24),  we  see  that 

it    crosses    the    axis    of    x    at    the    points 

B{-2,  0),   0(0,  0),  and   C(3,  0).    Hence 

part  of  the  area  is  above  the  axis  of  x  and 

part  below.     Accordingly,  we  shall  find  it 

necessary  to  solve  the  problem  in  two  parts, 

first  finding  the  area  above  the  axis  of  x 

and  then  finding  that  below.    To  find  the 

first  area  we  proceed  as  in  §  22,  dividing 

the  area  up  into  elementary  rectangles  for 

each  of  which 

dA  =  ydx  —  (x^  ~  x"^  —  Q  x)  dx ; 

whence  A  =  C\x^  -  x^  -  6  x)dx  =  [l  x*  -  ^  x^  -  3  x^f_^ 

=  0-11  (-  2/  -  ^  (-2)3  -  3  (-  2)2]  =  5.V 
Similarly,  for  the  area  below  the  axis  of  x  we  find,  as  before, 


Fig.  24 


dA  =  ydx  =  (x^ 


:)  dx. 


But  in  this  case  y  =  x^  —  x^  —  6  x  is  negative  and  hence  dA  is  negative, 
for  we  are  making  x  vary  from  0  to  3,  and  therefore  dx  is  positive.  There- 
fore we  expect  to  find  the  result  of  the  summation  negative.    In  fact. 

we  have  ^s 

A=       (x^-x^-6x)  dx  =  [lx*-  lx^-3  x"^-]^ 

=  [i  (3)^  -  U3)^  -  3  (3)2]  -  0  =  -  15^ 


DEFINITE  INTEGRAL 


65 


As  we  are  asked  to  compute  the  total  area  bounded  by  the  curve  and 
the  axis  of  x,  we  discard  the  negative  sign  in  the  last  summation  and  add 
5^  and  15 1,  thus  obtaining  21  ^^^  as  the  required  result. 

If  we  had  computed  the  definite  integral 

J- 2 

we  should  have  obtained  the  result  — 10  ,\,  which  is  the  algebraic  sum  of 
the  two  portions  of  area  computed  separately. 

Ex.  3.    Find  the  area  bounded  by  the  two  curves  y  =  x'^  and  y  =  8  —  x^. 
We  draw  the  curves  (Fig.  25) 

y  =  x^  (1) 

and  y  =  8  -  x^,  (2) 

and  by  solving  their  equations  we 
find  that  they  intersect  at  the  points 
Pi  (2,  2)  and  P^C- 2,  2). 

The  required  area  OP^BP^O  is  evi- 
dently twice  the  area  OP^BO,  since 
both  curves  are  symmetrical  with  re- 
spect to  OY.  Accordingly,  we  shall 
find  the  area  OP^BO  and  multiply  it 
by  2.  This  latter  area  may  be  found 
by  subtracting  the  area  ON^P^O  from 
the  area  ON^P^BO,  each  of  these  areas 
being  found  as  in  the  previous  example ; 
or  we  may  proceed  as  follows : 

Divide  ON.^  into  n  parts  dx,  and 
through  the  points  of  division  draw 
straight  lines  parallel  to  OY,  intersect- 
ing both  curves.  Let  one  of  these  lines 
be  MiQ^Q^.  Through  the  points  Qi  and 
Q2  draw  straight  lines  parallel  to  OX 

until  they  meet  the  next  vertical  line  to  the  right,  thereby  forming 
the  rectangle  Q^RSQ^.  The  area  of  such  a  rectangle  may  be  taken  as 
<//!  and  may  be  computed  as  follows :  its  base  is  dx  and  its  altitude  is 
Q1Q2  =  ^^A  -  ^^iQi  =  (8  -  a:2)  -  a:2  =  8  -  2  a:2 ;  for  Af^Q^  is  the  ordinate 
of  a  point  on  the  curve  (2)  and  M^Q^  the  ordinate  of  a  point  on  (1). 

Therefore  ,  .       ,0      o    2\  7 

dA  =  (8  —  2x^)dx', 

whence  A=  f^(8-2  x^)  dx  =  [8x-^x^l 

Jo  • 


Fig.  25 


=  [16 -J- (16)] 
Finally,  the  required  area  is  2(10§)  =  21^. 


0=10§. 


66  SUMMATION 

EXERCISES 

1.  Find  the  area  bounded  by  the  curve  4  y  —  a;^  —  2  =  0,  the  axis 
of  X,  and  the  lines  x  =  —  2  and  x  =  2. 

2.  Find  the  area  bounded  by  the  curve  y  =  x^  —  7  x^  -^  S  x  -\- 16, 
the  axis  of  x,  and  the  lines  x  =  1  and  x  =  3. 

3.  Find  the  area  bounded  by  the  curve  ?/  =  25  ic  —  10  x^  +  ic^  and 
the  axis  of  x. 

4.  Find   the   area   bounded  by  the   axis   of  x  and  the  curve 
y  =  25-  x\ 

5.  Find  the  area  bounded  by  the  curve  ?/  =  4£c^  —  4ic  —  3  and 
the  axis  of  x. 

6.  Find  the  area  bounded  below  by  the  axis  of  x  and  above  by 
the  curve  p  =  x^  —  4:  x^  —  A  x  -\-  16. 

7.  Find  the  area  bounded  by  the  curve  y  =  4: x^  —  S x^  —  9 x  -]-lS 
and  the  axis  oi  x. 

8.  Find  the  area  bounded  by  the  curve  x^  +  2?/— 8  =  0  and  the 
straight  line  x-\-2y— 6  =  0. 

9.  Find  the  area  bounded  by  the  curve  3y  —  x"^  =  0  and  the 
straight  line  2  x  -{-  y  —  9  =  0. 

10.  Find  the  area  of  the  crescent-shaped  figure  bounded  by  the 
two  curves  y  =  x"^  -\-7  and  y  =  2  x^  -\-  3. 

11.  Find   the   area   bounded  by  the  curves  4?/  =  cc^— 4cc  and 
x"- Ax -\-4.y- 24  =  0. 

12.  Find  the  area  bounded  by  the  curve  x-\-3  =  y'^—2y  and 
the  axis  of  y. 

24.  The  general  summation  problem.    The  formula 

'f(x)dx  =  F(h)-F{a)  (1) 


X 


has  been  obtained  by  the  study  of  an  area,  but  it  may  be  given 
a  much  more  general  application.  For  if  f(x)  is  any  function 
of  X  whatever,  it  may  be  graphically  represented  by  the  curve 
y  =zfQc).  The  rectangles  of  Fig.  23  are  then  the  graphical  rep- 
resentations of  the  products  f(x)  dx,  and  the  symbol  /  f{x)  dx 


GENERAL  PROBLEM  67 

represents  the  limit  of  the  sum  of  these  products.  We  may 
accordingly  say : 

Any  problem  which  requires  the  determination  of  the  limit  of  the 
sum  of  products  of  the  type  f(x)  dx  may  he  solved  hy  the  use  of 
formula  (1). 

Let  us  illustrate  this  by  considering  again  the  problem,  already 
solved  in  §  18,  of  determining  the  distance  traveled  in  the  time 
from  t  =  t^  to  t  =  t^hj  2i,  body  whose  velocity  v  is  known.    Since 

ds 

dt 
we  have  ds  =  vdt, 

which  is  approximately  the  distance  traveled  in  a  small  interval 
of  time  dt  Let  the  whole  time  from  ^  =  ^^  to  ^  =  ^^  ^  divided 
into  a  number  of  intervals  each  equal  to  dt  Then  the  total  dis- 
tance traveled  is  equal  to  the  sum  of  the  distances  traveled  in 
the  several  intervals  dt,  and  hence  is  equal  approximately  to  the 
sum  of  the  several  terms  vdt.  This  approximation  becomes  better 
as  the  size  of  the  intervals  dt  becomes  smaller  and  their  number 
larger,  and  we  conclude  that  the  limit  of  the  sum  of  the  terms  vdt 
is  the  actual  distance  traveled  by  the  body.  Hence  we  have,  if  s 
is  the  total  distance  traveled, 

vdt 


-I 


If,  now,  we  know  v  in  terms  of  f,  we  may  apply  formula  (1). 

Ex.  If  V  =  16  /  +  5,  find  the  distance  traveled  in  the  time  from  t  =  2  to 
<  =  4. 

We  have  directly 

s  =  C\lQ  I  +  5)  dt  =  [8  <2  4.  5  ^j4  ^  106. 

EXERCISES 

1.  At  any  time  t  the  velocity  of  a  moving  body  is  St^-\-2t  ft.  per 
second.   How  far  will  it  move  in  the  first  5  sec.  ? 

2.  How  far  will  the  body  in  Ex.  1  move  during  the  seventh  second  ? 

3.  At  any  time  t  the  velocity  of  a  moving  body  is  6  +  5  ^  —  ^^  ft.  per 
second.  Show  that  this  velocity  is  positive  during  the  interval  from 
t  =  —  l  tot  =  6,  and  find  how  far  the  body  moves  during  that  interval. 


68  SUMMATION 

4.  At  any  time  t  the  velocity  of  a  moving  body  is  4 ^2—  24^  -f  11  ft. 
per  second.  During  what  interval  of  time  is  the  velocity  negative,  and 
how  far  will  the  body  move  during  that  interval  ? 

5 .  The  number  of  foot  pounds  of  work  done  in  lifting  a  weight  is 
the  product  of  the  weight  in  pounds  and  the  distance  in  feet  through 
which  the  .weight  is  lifted.  A  cubic  foot  of  water  weighs  62^  lb. 
Compute  the  work  done  in  emptying  a  cylindrical  tank  of  depth  8  ft. 
and  radius  2  ft.,  considering  it  as  the  limit  of  the  sum  of  the  pieces 
of  work  done  in  lifting  each  thin  layer  of  water  to  the  top  of  the  tank. 

25.  Pressure.  It  is  shown  in  physics  that  the  pressure  on 
one  side  of  a  plane  surface  of  area  A,  immersed  in  a  liquid  at  a 
uniform  depth  of  h  units  below  the  surface  of  the  liquid,  is  equal 
to  wh A,  where  w  is  the  weight  of  a  unit  volume  of  the  liquid. 
This  may  be  remembered  by  noticing 
that  whA  is  the  weight  of  the  column 
of  the  liquid  which  would  be  supported 
by  the  area  A, 

Since  the  pressure  is  the  same  in 
all  directions,  we  can  also  determine  ^        -p^^  26 

the  pressure  on  one  side  of  a  plane 
surface  which  is  perpendicular  to  the  surface  of  the  liquid  and 
hence  is  not  at  a  uniform  depth. 

Let  ABC  (Fig.  26)  represent  such  a  surface  and  RS  the  line  of 
intersection  of  the  plane  of  ABC  with  the  surface  of  the  liquid. 
Divide  ABC  into  strips  by  drawing  straight  lines  parallel  to  RS. 
Call  the  area  of  one  of  these  strips  dA^  as  in  §  23,  and  the  depth  of 
one  edge  h.  Then,  since  the  strip  is  narrow  and  horizontal,  the 
depth  of  every  point  differs  only  slightly  from  A,  and  the  pres- 
sure on  the  strip  is  then  approximately  whdA,  Taking  P  as  the 
total  pressure,  we  write    ^p^^^^^ 

The  total  pressure  P  is  the  sum  of  the  pressures  on  the  several 

strips  and  is  therefore  the  limit  of  the   sum  of  terms  of  the 

form  whdAj  the  limit  being  approached  as  the  number  of  the 

strips  is  indefinitely  increased  and  the  width  of  each  indefinitely 

decreased.    Therefore  /> 

P=  j  whdA, 


PRESSURE  69 

where  the  limits  of  integration  are  to  be  taken  so  as  to  include 
the  whole  area  the  pressure  on  which  is  to  be  determined.  To 
evaluate  the  integral  it  is  necessary  to  express  both  h  and  dA 
in  terms  of  the  same  variable. 

Ex.  1.  Find  the  pressure  on  one  side  of  a  rectangle  BCDE  (Fig.  27), 
where  the  sides  BC  and  ED  are  each  4  ft.  long,  the  sides  BE  and  CD  are 
each  3  ft.  long,  immersed  in  water  so  that  the  plane  of  the  rectangle  is 
perpendicular  to  the  surface  of  the 
water,  and  the  side  ^C  is  parallel 
to  the  surface  of  the  water  and  2  ft. 
below  it. 

In  Fig.  27,  LK  is  the  line  of  inter- 
section of  the  surface  of  the  water 


0 ^ ^ 


Fig.  27 
X 


and  the  plane  of  the  rectangle.    Let  ~  M 

0  be  the  point  of  intersection  of  ^ 

LK  and  BE  produced.    Then,  if  a:  is 
measured  downward  from  0  along  E 

BE,  X  has  the  value  2  at  the  poi^t  B 
axid  X  has  the  value  5  at  the  point  E. 

We  now  divide  BE  into  parts  dx,  and  through  the  points  of  division 
draw  straight  lines  parallel  to  BC,  thus  dividing  the  given  rectangle 
into  elementary  rectangles  such  as  MNRS. 

Therefore        dA  =  area  of  MNRS  =  MN  •  M^  =  4  dx. 

Since  MN  is  at  a  distance  x  below  LK,  the  pressure  on  the  elementary 
rectangle  MNRS  is  approximately  wx(i:  dx).   Accordingly,  we  have 

dP  =  4  wxdx 

and  P=   f^iwxdx  =  [2icx^f=2w(dy-2w(2y  =  i2w. 

For  water,  w  =  62 1  lb.  =  ^l  t. 

Hence  we  have  finally 

.     P  =  2625  1b.  =  1/^T. 

Ex.  2.  The  base  CD  (Fig.  28)  of  a  triangle  BCD  is  7  ft.,  and  its  altitude 
from  B  to  CD  is  5  ft.  This  triangle  is  immersed  in  water  with  its  plane 
perpendicular  to  the  surface  of  the  water  and  with  CD  parallel  to  the  sur- 
face and  1  ft.  below  it,  B  being  below  CD.  Find  the  total  pressure  on  one 
side  of  this  triangle. 

Let  LK  represent  the  line  of  intersection  of  the  plane  of  the  triangle 
and  the  surface  of  the  water.  Then  B  is  6  ft.  below  LK.  Let  BX  be  per- 
pendicular to  LK  and  intersect  CD  at  T.  We  will  measure  distances 
from  B  in  the  direction  BX  and  denote  them  by  x.  Then,  at  the  point  B, 
X  has  the  value  0 ;  and  at  T,  x  has  the  value  5. 


70 


SUMMATION 


Divide  the  distance  BT  into  parts  dx,  and  through  the  points  of  divi- 
sion draw  straight  lines  parallel  to  CD,  and  on  each  of  these  lines  as 
lower  base  construct  a  rectangle  such  as  MNRS,  where  E  and  F  are 
two  consecutive  points  of  division  -^ 

on  BX. 

Then       BE  =  x, 
EF  =  dx, 
and,  by  similar  triangles, 

MN  ^  BE  ^ 
CD  ~  BT' 

MN      X 

whence      -— —  =  - 

7         5 

and  MN  =lx. 

Then 

Since  £  is  6  ft.  below  LK,  and  BE 
below  LK. 

Hence  the  pressure  on  the  rectangle  is  approximately 

dP  =  (Ix dx)  (6  —  x)w  =  (*/-  wx  —  I  icx^)  dx, 

and  ^=  f  VV  wx-  I  tvx^)  dx  =  [V"  w?^^  -  75  w^^]^ 

Jo 
=  (105iv  -  ^^-  tv)  -  0  =  i|o_  ,^  :^  2916§  lb.  =  l|i  T. 


dA  =  the  area  of  MNRS  =  J  xdx. 


it  follows  that  ^  is  (6  -  x)  ft. 


EXERCISES 

1.  A  gate  in  the  side  of  a  dam  is  in  the  form  of  a  square,  4  ft. 
on  a  side,  the  upper  side  being  parallel  to  and  15  ft.  below  the  surface 
of  the  water  in  the  reservoir.    What  is  the  pressure  on  the  gate  ? 

2.  Find  the  total  pressure  on  one  side  of  a  triangle  of  base  6  ft. 
and  altitude  6  ft.,  submerged  in  water  so  that  the  altitude  is  vertical 
and  the  vertex  is  in  the  surface  of  the  water. 

3.  Find  the  total  pressure  on  one  side  of  a  triangle  of  base  4  ft. 
and  altitude  6  ft.,  submerged  in  water  so  that  the  base  is  horizontal, 
the  altitude  vertical,  and  the  vertex  above  the  base  and  4  ft.  from  the 
surface  of  the  water. 

4.  The  base  of  an  isosceles  triangle  is  8  ft.  and  the  equal  sides 
are  each  5  ft.  The  triangle  is  completely  immersed  in  water,  its  base 
being  parallel  to  and  5  ft.  below  the  surface  of  the  water,  its  alti- 
tude being  perpendicular  to  the  surface  of  the  water,  and  its  vertex 
being  above  the  base.  Find  the  total  pressure  on  one  side  of  the 
triangle. 


VOLUME  71 

5.  Find  the  pressure  on  one  side  of  an  equilateral  triangle,  6  ft. 
on  a  side,  if  it  is  partly  submerged  in  liquid  so  that  one  vertex  is 
one  foot  above  the  surface  of  the  liquid,  the  corresponding  altitude 
being  perpendicular  to  the  surface  of  the  liquid. 

6.  The  gate  in  Ex.  1  is  strengthened  by  a  brace  which  runs 
diagonally  from  one  corner  to  another.  Find  the  pressure  on"~each — 
of  the  two  portions  of  the  gate  —  one  above,  the  other  below,  the 
brace. 

7.  A  dam  is  in  the  form  of  a  trapezoid,  with  its  two  horizontal 
sides  300  and  100  ft.  respectively,  the  longer  side  being  at  the  top ; 
and  the  height  is  15  ft.  What  is  the  pressure  on  the  dam  when  the 
water  is  level  with  the  top  of  the  dam  ? 

8.  What  is  the  pressure  on  the  dam  of  Ex.  7  when  the  water 
reaches  halfway  to  the  top  of  the  dam  ? 

9.  If  it  had  been  necessary  to  construct  the  dam  of  Ex.  7  with 
the  shorter  side  at  the  top  instead  of  the  longer  side,  how  much 
greater  pressure  would  the  dam  have  had  to  sustain  when  the 
reservoir  is  full  of  water  ? 

10.  The  center  board  of  a  yacht  is  in  the  form  of  a  trapezoid  in 
which  the  two  parallel  sides  are  3  ft.  and  6  ft.,  respectively,  in  length, 
and  the  side  perpendicular  to  these  two  is  4  ft.  in  length.  Assuming 
that  the  last-named  side  is  parallel  to  the  surface  of  the  water  at 
a  depth  of  2  ft.,  and  that  the  parallel  sides  are  vertical,  find  the 
pressure  on  one  side  of  the  board. 

11.  Where  shall  a  horizontal  line  be  drawn  across  the  gate  of 
Ex.  1  so  that  the  pressure  on  the  portion  above  the  line  shall  equal 
the  pressure  on  the  portion  below  ? 

26.  Volume.  The  volume  of  a  solid  may  be  computed  by  di- 
viding it  into  n  elements  of  volume,  dV,  and  taking  the  limit 
of  the  sum  of  these  elements  as  n  is  increased  indefinitely,  the 
magnitude  of  each  element  at  the  same  time  approaching  zero. 
The  question  in  each  case  is  the  determination  of  the  form  of 
the  element  dV.  We  shall  discuss  a  comparatively  simple  case 
of  a  solid  such  as  is  shown  in  Fig.  29. 

In  this  figure  let  OH  be  a  straight  line,  and  let  the  distance 
of  any  point  of  it  from  0  be  denoted  by  h.  At  one  end  the  solid 
is  bounded  by  a  plane  perpendicular  to  OH  dX  C,  where  OC=a^ 


72 


SUMMATION 


and  at  the  other  end  it  is  bounded  by  a  plane  perpendicular  to 
OH  at  J5,  where  OB  =  5,  so  that  it  has  parallel  bases. 

The  solid  is  assumed  to  be  such  that  the  area  A  of  any  plane 
section  made  by  a  plane  perpendicular  to  OH  at  a  point  distant 
h  from  0  can  be  expressed  as  a  func- 
tion of  h. 

To  find  the  volume  of  such  a  solid 
we  divide  the  distance  CB  into  n  parts 
dh,  and  through  the  points  of  division 
pass  planes  perpendicular  to  OH.  We 
have  thus  divided  the  solid  into  slices 
of  which  the  thickness  is  dh. 

Since  A  is  the  area  of  the  base  of  a 
slice,  and  since  the  volume  of  the  slice 
is  approximately  equal  to  the  volume 
of  a  right  cylinder  of  the  same  base 
and  thickness,  we  write 

dV=Adk 

The  volume  of  the  solid  is  then  the  limit  of  the  sum  of  terms  of 
the  above  type,  and  therefore 

V=  C  Adh. 

It  is  clear  that  the  above  discussion  is  valid  even  when  one 
or  both  of  the  bases  corresponding  ioh  =  a  and  h  =  h,  respectively, 
reduces  to  a  point. 

Ex.1.  Let  OF  (Fig.  30) 
be  an  edge  of  a  solid  such 
that  all  its  sections  made 
by  planes  perpendicular  to 
OF  are  rectangles,  the  sides 
of  a  rectangle  in  a  plane 
distant  y  from  O  being  re- 
spectively 2  y  and  y\  We 
shall  find  the  volume  in- 
cluded between  the  planes 
y  =  0  and  3/ =  2^. 

Dividing  the  distance  from  y  =  Q  io  y  =  2\  into  n  parts  dy,  and  passing 
planes  perpendicular  to  OY,  we  form  rectangles  such  as  MNRS,  where,  if 


VOLUME  73 

OM  =  ij,  MN  =  3/2  and  MS  =  2  y.    Hence  the  area  MNRS  =  2y\  and  the 
volume  of  the  elementary  cylinder  standing  on  MNRS  as  a  base  is  2y^dy\ 

*^^*'^'  dV=2fdy. 

Therefore  V  =  C^^y^dy  =  [^  y^f^=  19^|. 

Ex.  2.  The  axes  of  two  equal  right  circular  cylinders  of  radius  a  inter- 
sect at  right  angles.    Required  the  volume  common  to  the  two  cylinders. 

Let  OA  and  OB  (Fig.  31)  be  the  axes  of  the 
cylinders  and  OY  the  common  perpendicular  to  i 

OA  and  OB  at  their  point  of  intersection  0.   Then  ^| 

OAD  and  OBD  are  quadrants  of  two  equal  circles 
cut  from  the  two  cylinders  by  the  planes  through  / 

OY  perpendicular  to  the  axes  OB  and  OA,  and  / 

OD  =  a.  Then  the  figure  represents  one  eighth     -pL"'' 
of  the  required  volume.  /^ 

We  divide  the  distance  OB  into  n  parts  dy,      /     ,^^^^     \     ^./^"^ 

and  through  the  points  of  division  pass  planes  per-     [yi V^ 

pendicular  to  OY.   Any  section,  such  as  LMNP^    ^  ^ 

is  a  square,  of  which  one  side  NP  is  equal  to  Fig.  31 

WOP^—  ON^.    OP  =  a,  being  a  radius  of  one  of  the  cylinders,  and  hence, 
as  0N=  y,  

NP  =  Va2  -  /. 

Accordingly,  the  area  of  LMNP  =  a^  —  y^,  and  the  volume  of  the  ele- 
mentary cylinder  standing  on  LMNP  as  a  base  is 

dV=  (a'^-y'^)dy; 

whence  V  =  J \a^  -  y^)  dy  =  [a^y  -  ^  y^]-=  i  a^ 

Hence  the  total  volume  is  ^^  a'. 

This  method  of  finding  volumes  is  particularly  useful  when 
the  sections  of  the  solid  made  by  parallel  planes  are  bounded 
by  circles  or  by  concentric  circles.  Such  a  solid  may  be  gen- 
erated by  the  revolution  of  a  plane  area  around  an  axis  in  its 
plane,  and  is  called  a  solid  of  revolution.  We  take  the  following 
examples  of  solids  of  revolution  : 

Ex.  3.  Find  the  volume  of  the  solid  generated  by  revolving  about  OX 
the  area  bounded  by  the  curve  y^  =  4:Xf  the  axis  of  x,  and  the  line  x  =  3. 

The  generating  area  is  shown  in  Fig.  32,  where  ^  i5  is  the  line  x  =  3. 
Hence  OA  =  3. 


74 


SUMMATION 


Divide  OA  into  n  parts  dx,  and  through  the  points  of  division  pass 
straight  lines  parallel  to  OY,  meeting  the  curve.  When  the  area  is  revolved 
about  OX,  each  of  these  lines,  as  AIP,  NQ,  etc.,  generates  a  circle,  the  plane 
of  which  is  perpendicular  to  OX.  The  area 
of  the  circle  generated  by  MP,  for  example, 
is  ttMP  ^,  which  is  equal  to  tt?/^  =  tt  (4  x),  if 
OM=x. 

Hence  the  area  of  any  plane  section  of 
the  solid  made  by  a  plane  perpendicular  to 
OX  can  be  expressed  in  terms  of  its  dis- 
tance from  0,  and  we  may  apply  the  previous 
method  for  finding  the  volume. 

Since  the  base  of  any  elementary  cylinder 
is  4  TTX  and  its  altitude  is  dx,  we  have 
dV  =  4:Trx  dx. 

Hence    V=  C^^Trxdx  =  [2  7rx^f^  =  lSTr.  ^^^   32 

Ex.  4.    Find  the  volume  of  the  ring  surface  generated  by  revolving  about 
the  axis  of  x  the  area  bounded  by  the  line  y  =  5  and  the  curve  y  =  Q  —  x^. 

The  line  and  the  curve  (Fig.  33)  are 
seen  to  intersect  at  the  points  Pi(—  2,  5) 
and  P^i^,  5),  and  the  ring  is  generated 
by  the  area  P^BP^P^.  Since  this  area  is 
symmetrical  with  respect  to  OY,  it  is  evi- 
dent that  the  volume  of  the  ring  is  twice 
the  volume  generated  by  the  area  AP^BA. 
Accordingly,  we  shall  find  the  latter  volume 
and  multiply  it  by  2. 

We  divide  the  line  OM^  =  2  (M^  being 
the  projection  of  P^  on  OX)  into  n  parts 
dx,  and  through  the  points  of  division  draw 
straight  lines  parallel  to  OF  and  intersect- 
ing the  straight  line  and  the  curve.  One 
of  these  lines,  as  MQP,  will,  when  revolved 
about  OX,  generate  a  circular  ring,  the 
outer  radius  of  which  is  MP  =  y  =  9  —  x^ 
and  the  inner  radius  of  which  is  MQ=y  =  5. 
Hence  the  area  of  the  ring  is 

ttMP^-  ttMQ^  =  7r(9  -  x^  -  7r(5)2 

=  '7r(56-18a:2-f24).  Fig.  33 

Accordingly,  'dV  =  v  (oQ  —  18  x^  +  x^)  dx 

and       V=  f^Tr(5Q  -18  x^  ■{■  x^)dx  =  7r[5Q  X  -  Q  x^  +  la:^]^  =  70|ir. 

t/O 

Accordingly,  the  volume  of  the  ring  is  2  (70 §  tt)  =  140|  ir. 


VOLUME  75 

EXERCISES 

1.  The  section  of  a  certain  solid  made  by  any  plane  perpendicular 
to  a  given  line  OH  is  a  circle  with  one  point  in  OH  and  its  center 
on  a  straight  line  intersecting  OH  at  an  angle  of  45°.  If  the  height 
of  this  solid  measured  from  0  along  OH  is  4  ft.,  find  its  volume  by 
integration. 

2.  A  solid  is  such  that  any  cross  section  perpendicular  to  an 
axis  is  an  equilateral  triangle  of  which  each  side  is  equal  to  the 
square  of  the  distance  of  the  plane  of  the  triangle  from  a  fixed  point 
on  the  axis.  The  total  length  of  the  axis  from  the  fixed  point  is  5. 
Find  the  volume. 

3.  Find  the  volume  of  the  solid  generated  by  revolving  about  OX 
the  area  bounded  by  OX  and  the  curve  y  =  A:X  —  x^. 

4.  Find  the  volume  of  the  solid  generated  by  revolving  about  OX 
the  area  included  between  the  axis  of  x  and  the  curve  y  =  2x^  —  x^. 

5.  Find  the  volume  of  the  solid  generated  by  revolving  about  the 
line  ?/  =  —  2  the  area  bounded  by  the  axis  of  y,  the  lines  a:  =  3 
and  ?/  =  —  2,  and  the  curve  y  =  3x^. 

6.  On  a  spherical  ball  of  radius  5  in.  two  great  circles  are  drawn 
intersecting  at  right  angles  at  the  points  A  and  B.  The  material  of 
the  ball  is  then  cut  away  so  that  the  sections  perpendicular  to  ^jB 
are  squares  with  their  vertices  on  the  two  great  circles.  Find  the 
volume  left. 

7.  Find  the  volume  generated  by  revolving  about  the  line  x  =  2 
the  area  bounded  by  the  curve  y^z=  Sx,  the  axis  of  x,  and  the 
line  X  =  2. 

8.  Any  plane  section  of  a  certain  solid  made  by  a  plane  perpen- 
dicular to  OF  is  a  square  of  which  the  center  lies  on  OY  and  two 
opposite  vertices  lie  on  the  curve  y  =  4:X^.  Find  the  volume  of  the 
solid  if  the  extreme  distance  along  OY  is  3. 

9.  Find  the  volume  generated  by  revolving  about  *0Y  the  area 
bounded  by  the  curve  y^=  Sx  and  the  line  x  =  2, 

10.  Find  the  volume  of  the  solid  generated  by  revolving  about  OX 
the  area  bounded  by  the  curves  y  =  6x  —  x^  and  y  =  x^  —  6x  -\-10. 

11.  A  variable  square  moves  with  its  plane  perpendicular  to  OX, 
so  that  the  ends  of  one  of  its  sides  are  on  the  curves  16y  =  x"  and 
4:y  =  x^  —  12.  Find  the  volume  of  the  solid  generated  as  the  square 
takes  all  possible  positions  under  the  given  conditions. 


76  SUMMATION 

GENERAL  EXERCISES 

1.  The  velocity  in  feet  per  second  of  a  moving  body  at  any 
time  t  is  ^^—4^  +  4.  Show  that  the  body  is  always  moving  in 
the  direction  in  which  s  is  measured,  and  find  how  far  it  will  move 
during  the  fifth  second. 

2.  The  velocity  in  feet  per  second  of  a  moving  body  at  any  time 
^  is  ^^  —  4  ^.  Show  that  after  j^  =  4  the  body  will  always  move  in 
the  direction  in  which  s  is  measured,  and  find  how  far  it  will  move 
in  the  time  from  ^  =  6  to  ^  =  9. 

3.  At  any  time  t  the  velocity  in  feet  per  second  of  a  moving 
body  is  ^^—6^  +  5.  How  many  feet  will  the  body  move  in  the 
direction  opposite  to  that  in  which  s  is  measured  ? 

4.  At  any  time  t  the  velocity  in  miles  per  hour  of  a  moving  body 
is  ^^  —  2  ^  —  3.  If  the  initial  moment  of  time  is  12  o'clock  noon,  how 
far  will  the  body  move  in  the  time  from  11.30  a.m.  to  2  p.m.  ? 

5 .  Find  the  area  bounded  by  the  curves  9y  =  ^x^  and  45  —  9 ?/  =  cc^. 

6.  Find  the  total  area  bounded  by  the  curves  y^=  ^ax  and 
7/^  =  4  a^  —  4  ax. 

7.  Find  the  total  area  bounded  by  the  curve  y  =  x^  and  the 
straight  line  y  =  A:X. 

8.  Find  the  total  area  bounded  by  the  curve  y  —  x(x  —  l)(x  —  ^) 
and  the  straight  line  ?/  =  4  (x  —  1). 

9.  ABCD  is  a  quadrilateral  with  A  =  90°,  B  =  90°,  ^^  =  5  ft., 
EC  =  2  ft.,  ^  Z>  =  4  f t.  It  is  completely  immersed  in  water  with  AB 
in  the  surface  and  AD  and  BC  perpendicular  to  the  surface.  Find 
the  pressure  on  one  side. 

10.  Prove  that  the  pressure  on  one  side  of  a  rectangle  completely 
submerged  with  its  plane  vertical  is  equal  to  the  area  of  the  rectangle 
multiplied  by  the  depth  of  its  center  and  by  w  (consider  only  the 
case  in  which  one  side  of  the  rectangle  is  parallel  to  the  surface). 

11.  Prove  that  the  pressure  on  one  side  of  a  triangle  completely 
submerged  with  its  plane  vertical  is  equal  to  its  area  multiplied  by 
the  depth  of  its  median  point  and  by  w  (consider  only  the  case  in 
which  one  side  of  the  triangle  is  parallel  to  the  surface). 

12.  The  end  of  a  trough,  full  of  water,  is  assumed  to  be  in  the 
form  of  an  equilateral  triangle,  with  its  vertex  down  and  its  plane 
vertical.  What  is  the  effect  upon  the  pressure  on  the  end  if  the 
level  of  the  water  sinks  halfway  to  the  bottom  ? 


GENERAL  EXERCISES  77 

13.  A  square  2  ft.  on  a  side  is  immersed  in  water,  with  one  vertex 
in  the  surface  of  the  water  and  with  the  diagonal  through  that 
vertex  perpendicular  to  the  surface  of  the  water.  How  much  greater 
is  the  pressure  on  the  lower  half  of  the  square  than  that  on  the 
upper  half? 

14.  A  board  is  symmetrical  with  respect  to  the  line  AB^  and  is  of 
such  a  shape  that  the  length  of  any  line  across  the  board  perpendic- 
ular to  AB  is  twice  the  cube  of  the  distance  of  the  line  from  A. 
^-B  is  2  ft.  long.  The  board  is  totally  submerged  in  water,  AB  being 
perpendicular  to  the  surface  of  the  water  and  A  one  foot  below  the 
surface.    Find  the  pressure  on  one  side  of  the  board. 

15.  Find  the  pressure  on  one  side  of  an  area  the  equations  of  whose 
boundary  lines  are  ic  =  0,  i/  =  4,  and  y^  =  4:X  respectively,  where  the 
axis  of  X  is  taken  in  the  surface  of  the  water  and  where  the  positive 
direction  of  the  y  axis  is  downward  and  vertical. 

16.  Find  the  volume  generated  by  revolving  about  OX  the  area 
bounded  by  OX  and  the  curve  4  y  =  16  —  ic^. 

17.  Find  the  volume  generated  by  revolving  about  OX  the  area 
bounded  by  the  curve  y  =  x^  +  2  and  the  line  y  =  S. 

18.  Find  the  volume  generated  by  revolving  about  OX  the  area 
bounded  by  OX  and  the  curve  y  =  S  x  —  x^. 

19.  Find  the  volume  generated  by  revolving  about  the  line  y  =  —  l 
the  area  bounded  by  the  curves  9y  =  2x^  and  9 y  =  S6  —  2 x'^. 

20.  An  axman  makes  a  wedge-shaped  cut  in  the  trunk  of  a  tree. 
Assuming  that  the  trunk  is  a  right  circular  cylinder  of  radius  8  in., 
that  the  lower  surface  of  the  cut  is  a  horizontal  plane,  and  that  the 
upper  surface  is  a  plane  inclined  at  an  angle  of  45°  to  the  horizontal 
and  intersecting  the  lower  surface  of  the  cut  in  a  diameter,  find  the 
amount  of  wood  cut  out. 

21.  On  a  system  of  parallel  chords  of  a  circle  of  radius  2  there 
are  constructed  equilateral  triangles  with  their  planes  perpendicular 
to  the  plane  of  the  circle  and  on  the  same  side  of  that  plane,  thus 
forming  a  solid.    Find  the  volume  of  the  solid. 

22.  Show  that  the  volume  of  the  solid  generated  by  revolving 
about  OY  the  area  bounded  by  OX  and  the  curve  y  =  a  —  bx^  is 
equal  to  the  area  of  the  base  of  the  solid  multiplied  by  half  its  altitude. 

23.  In  a  sphere  of  radius  a  find  the  volume  of  a  segment  of  one 
base  and  altitude  h. 


78  SUMMATION 

24.  A  solid  is  such  that  any  cross  section  perpendicular  to  an  axis 
is  a  circle,  with  its  radius  equal  to  the  square  root  of  the  distance  of 
the  section  from  a  fixed  point  of  the  axis.  The  total  length  of  the 
axis  from  the  fixed  point  is  4.    Find  the  volume  of  the  solid. 

25.  A  variable  square  moves  with  its  plane  perpendicular  to  the 
axis  of  y  and  with  the  ends  of  one  of  its  diagonals  respectively  in 
the  parts  of  the  curves  y^  =  lQx  and  y^  =  4  ic,  which  are  above 
the  axis  of  x.  Find  the  volume  generated  by  the  square  as  its  plane 
moves  a  distance  8  from  the  origin. 

26.  The  plane  of  a  variable  circle  moves  so  as  to  be  perpendicular 
to  OXj  and  the  ends  of  a  diameter  are  on  the  curves  y  =  x^  and 
y  =  Sx^  —  8.  Find  the  volume  of  the  solid  generated  as  the  plane 
moves  from  one  point  of  intersection  of  the  curves  to  the  other. 

27.  All  sections  of  a  certain  solid  made  by  planes  perpendicular 
to  OF  are  isosceles  triangles.  The  base  of  each  triangle  is  a  line 
drawn  perpendicular  to  0  Y,  with  its  ends  in  the  curve  y  =  Ax"^.  The 
altitude  of  each  triangle  is  equal  to  its  base.  Find  the  volume  of 
the  solid  included  between  the  planes  for  which  y  =  0  and  y  =  Q. 

28.  All  sections  of  a  certain  solid  made  by  planes  perpendicular 
to  OF  are  right  isosceles  triangles.  One  leg  of  each  triangle  coincides 
with  the  line  perpendicular  to  OF  with  its  ends  in  OF  and  the  curve 
y^  =  4:X.  Find  the  volume  of  the  solid  between  the  sections  for  which 
y  =  0  and  y  =  S. 

29.  Find  the  work  done  in  pumping  all  the  water  from  a  full 
cylindrical  tank,  of  height  15  ft.  and  radius  3  ft.,  to  a  height  of 
20  ft.  above  the  top  of  the  tank. 

30.  Find  the  work  done  in  emptying  of  water  a  full  conical 
receiver  of  altitude  6  ft.  and  radius  3  ft.,  the  vertex  of  the  cone 
being  down. 


CHAPTER  IV 


ALGEBRAIC  FUNCTIONS 

27.  Distance  between  two  points.  Let  ij  (x^,  j/^)  and  I^x^,  y^) 
(Fig.  34)  be  any  two  points  in  the  plane  XOF,  such  that  the 
straight  line  iJTJ  is  not  parallel  either  to  OX  or  to  0  Y,  Through 
ij  draw  a  straight  line  parallel 
to  OX,  and  through  i^  draw  a 
straight  line  parallel  to  OF,  and 
denote  their  point  of  intersection 
by  iJ. 

Then  P^R  =  A.r  =  x^—  x^ 
and         RP, 


In  the  right  triangle  PfiP^ 


p^p^  =  ^pX+rp^' 


^ 

p. 

/ 

0 

^ 

I 

I 

Fig.  34 


whence 


P,P,  =  y/(x,-x,y-^iy,-y,y 


•(1) 


If  y^^y^^  P^P^  is  parallel  to  OX,  and  the  formula  reduces  to 

P,P,  =  x^-x^.  (2) 

In  like  manner,  if  x^=x^^   P^P^  is  parallel  to   OF,  and  the 

p.p.  =  y.-y.'  (3) 


formula  reduces  to 


28.  Circle.  Since  a  circle  is  the  locus  of  a  point  which  is 
always  at  a  constant  distance  from  a  fixed  point,  formula  (1) 
§  27,  enables  us  to  write  down  immediately  the  equation  of  a 
circle. 

Let  0(A,  ^)  (Fig.  35)  be  the  center  of  a  circle  of  radius  r. 
Then,  if  P(x,  y)  is  any  point  of  the  circle,  by  (1),  §  27,  a:  and 
y  must  satisfy  the  equation 

Qc-hy+iy-hf^r'.  (1) 

79 


-X 


Fig.  35 


80  ALGEBRAIC  FUNCTIONS 

Moreover,  any  point  the  coordinates  of  which  satisfy  (1), 
must  be  at  the  distance  r  from  C  and  hence  be  a  point  of  the 
circle.    Accordingly,  (1)  is  the  equation  of  a  circle. 

If  (1)  is  expanded,  it  becomes 

x''-hf-2hx-2Jcf/-\-h'-\-k'-r'=0,  (2) 

an   equation    of  the   second   degree  with  no  term  in  xi/  and 

with  the  coefficients  of  x^  and  y^        „ 

equal. 

Conversely,  any  equation  of  the 
second  degree  with  no  xi/  term  and 
with  the  coefficients  of  x^  and  «/^ 
equal  (as 

Aa^-{-Af-{-2Gx-{-2Fi/-\-C=0,  (3)  _ 

where  A,  G,  F,  and  C  are  any  con- 
stants) may  be  transformed  into 
the  form  (1)  and  represents  a  circle,  unless  the  number  cor- 
responding to  r^  is  negative  (see  Ex.  3,  page  81),  in  which 
ease  the  equation  is  satisfied  by  no  real  values  of  x  and  y  and 
accordingly  has  no  corresponding  locus. 

The  circle  is  most  readily  drawn  by  making  such  transfor- 
mation, locating  the  center,  and  constructing  the  circle  with 
compasses. 

Ex.1.  x^  +  y^-2x-iy  =  0. 

This  equation  may  be  written  in  the  form 

(x^-2x       )4-(/-4y        )  =  0, 

and  the  terms  in  the  parentheses  may  be  made  perfect  squares  by  adding 
1  in  the  first  parenthesis  and  4  in  the  second  parenthesis.  As  we  have 
added  a  total  of  5  to  the  left-hand  side  of  the  equation,  we  must  add  an 
equal  amount  to  the  right-hand  side  of  the  equation.    The  result  is 

(3^2  -  2  a:  -f- 1)  -f  (y2  _  4  y  -1-  4)  =  5, 

which  may  be  placed  in  the  form 

(x  - 1)2  +  (y-  2)2  =  5, 

the  equation  of  a  circle  of  radius  Vo  with  its  center  at  the  point  (1,  2). 


CIRCLE  81 

Ex.  2.  9x^  +  9  f -Ox +  Q7j-S  =  0. 

Placing  8  on  the  right-hand  side  of  the  equation  and  then  dividing  by 
9,  we  have  212  .    ^  s 

which  may  be  treated  by  the  method  used  in  Ex.  1.    The  result  is 

the  equation  of  a  circle  of  radius  ^VE,  with  its  center  at  (i,  —J). 
Ex.3.  9^:2  +  9/ -6a; +12?/ +  11  =  0. 

Proceeding  as  in  Ex.  2,  we  have,  as  the  transformed  equation, 

an  equation  which  cannot  be  satisfied  by  any  real  values  of  x  and  y,  since 
the  sum  of  two  positive  quantities  cannot  be  negative.  Hence  this  equation 
corresponds  to  no  real  curve. 

EXERCISES 

1.  Find  the  equation  of  the  circle  with  the  center  (4,  —  2)  and 
the  radius  3. 

2.  Find  the  equation  of  the  circle  with  the  center  (0,  —  1)  and 
the  radius  5. 

3.  Find  the  center  and  the  radius  of  the  circle 

ic'  +  2/'+6ic-10?/  +  9  =  0. 

4.  Find  the  center  and  the  radius  of  the  circle 

5x'-\-5y^+Sx-6i/-15  =  0. 

5.  Find  the  equation  of  the  straight  line  passing  through  the 

center  of  the  circle 

x'  +  f+2x-y  +  l  =  0 

and  perpendicular  to  the  line 

2a;  +  3?/-4  =  0. 

6.  Prove  that  two  circles  are  concentric  if  their  equations  differ 
only  in  the  absolute  term. 

29.  Parabola.  The  locus  of  a  point  equally/  distant  from  a  fixed 
point  and  a  fixed  straight  line  is  called  a  parabola.  The  fixed 
point  is  called  the  focus  and  the  fixed  straight  line  is  called  the 
directrix. 


82  ALGEBRAIC  FUNCTIONS 

Let  F  (Fig.  36)  be  the  focus  and  ^*S^  the  directrix  of  a 
parabola.  Through  F  draw  a  straight  Hue  perpendicular  to 
RS^  intersecting  it  at  Z>,  and  let  this  line  be  the  axis  of  x. 
Let  the  middle  point  of  DF  be  taken  as  0,  the  origin  of  coordi- 
nates, and  draw  the  axis  OY.  Then,  if  the  distance  DF'i^  2  c,  the 
coordinates  of  i^  are  (c?,  0)  and  the  equation  of  RS  is  x=—c. 

Let  P(x^  y)  be   any  point   of  the   parabola,  and  draw  the 
straight  line  FP  and  the  straight  line  NF      SY 
perpendicular  to  RS. 

Then  NF  =  x-\-  c, 


N 

p^ 

(f 

D 

\F 

and,  by  §  27,  FF=  ^  {x  -  cf^-  if-,  ~^ ^ 

whence,  from  the  definition  of  the  parabola, 

^^^cf^f={x^c)\ 
which  reduces  to        i/=4:cx.  (1)    ^       Fig.  36 

Conversely,  if  the  coordinates  of  any  point  P  satisfy  (1),  it 
can  be  shown  that  the  distances  FP  and  NP  are  equal,  and 
hence  P  is  a  point  of  the  parabola. 

Solving  (1)  for  y  in  terms  of  x,  we  have 

^  =  ±2V^.  (2) 

We  assume  that  c  is  positive.  Then  it  is  evident  that  if  a 
negative  value  is  assigned  to  x^  y  is  imaginary,  and  no  correspond- 
ing points  of  the  parabola  can  be  located.  All  positive  values 
may  be  assigned  to  x^  however,  and  hence  the  parabola  lies 
entirely  on  the  positive  side  of  the  axis  OY. 

Accordingly,  we  assign  positive  values  to  a:,  compute  the  cor- 
responding values  of  y^  and  draw  a  smooth  curve  through  the 
points  thus  located. 

It  is  to  be  noticed  that  to  every  value  assigned  to  x  there  are 
two  corresponding  values  of  «/,  equal  in  magnitude  and  opposite 
in  algebraic  sign,  to  which  there  correspond  two  points  of  the 
parabola  on  opposite  sides  of  OX  and  equally  distant  from  it. 
Hence  the  parabola  is  symmetrical  with  respect  to  OX,  and  ac- 
cordingly OX  is  called  the  axis  of  the  parabola. 

The  point  at  which  its  axis  intersects  a  parabola  is  called  the  ver- 
tex of  the  parabola.   Accordingly,  0  is  the  vertex  of  the  parabola. 


PARABOLA  83 

Returning  to  Fig.  36,  if  F  is  taken  at  the  left  of  0  with  the 
coordinates  (—  c^  0),  and  RS  is  taken  at  the  right  of  0  with  the 
equation  a;  =  c?,  equation  (1)  becomes 

y''=-4:cx  (3) 

and  represents  a  parabola  lying  on  the  negative  side  of  OY. 
Hence  we  conclude  that  any  equation  in  the  form 

y''=lcx,  (4) 

where  A:  is  a  positive  or  a  negative  constant,  is  a  parabola,  with 

its  vertex  at  O,  its  axis  on  OX,  its  focus  at  the  point  (-»  0), 

and  its  directrix  the  straight  line  x  =  ~  -• 

4 

Similarly,  the  equation       x^='ky        '  (5) 

represents  a  parabola,  with  its  vertex  at  0  and  with  its  axis  coin- 
ciding with  the  positive  or  the  negative  part  of  OF,  according 
as  Jc  is  positive  or  negative.    The  focus  is  always  the  point 

/     k\  k  . 

(0,  -h  and  the  directrix  is  the  line  ?/  =  —-,  whether  k  be  positive 

or  negative. 

30.  Parabolic  segment.  An  important  property  of  the  parabola 
is  contained  in  the  following  theorem: 

The  square  of  any  two  chords  of  a  parabola  which  are  perpen- 
dicular to  its  axis  are  to  each  other  as  their  distances  from  the 
vertex  of  the  parabola. 

This  theorem  may  be  proved  as  follows: 
Let  F^(x^,  ?/i)  and  P^Cx^^  y.2^  be  any  two  points  of  any  parab- 
ola y''=kx  (Fig.  37). 

Then  y^  =  kx^ 

and  y2  =  kx^\ 

whence 

whence  11^  =  :^.  (1) 


yf 

=  ?!, 

(2yJ' 

^2 

84 


ALGEBRAIC  FUNCTIONS 


From  the  symmetry  of  the  parabola,  2?/^=  Q^P^  and  2^/^=  Q^I^. 


But  X, 


OM^  and  x  =  OM^ 


and  hence 


(1)  becomes 


and  the  theorem  is  proved. 

The  figure  bounded  by  the  parabola 
and  a  chord  perpendicular  to  the  axis 
of  the  parabola,  as  QfiP^  (Fig.  37), 
is  called  a  'parabolic  segment.  The 
chord  is  called  the  ha%e  of  the  segment,  the  vertex  of  the 
parabola  is  called  the  vertex  of  the  segment,  and  the  distance 
from  the  vertex  to  the  base  is  called  the  altitude  of  the  segment. 


Fig.  37 


EXERCISES 


Plot  the  following  parabolas,  determining  the  focus  of  each : 
1.  if  =  —  Sec.  3.  2/2=  ^x. 


2.  x' 


42/. 


4.  x^ 


7  2/. 


5.  The  altitude  of  a  parabolic  segment  is  10  ft.,  and  the  length  of 
its  base  is  16  ft.  A  straight  line  drawn  across  the  segment  perpen- 
dicular to  its  axis  is  10  ft.  long.  How  far  is  it  from  the  vertex  of 
the  segment  ? 

6.  An  arch  in  the  form  of  a  parabolic  curve,  the  axis  being 
vertical,  is  50  ft.  across  the  bottom,  and  the  highest  point  is  15  ft. 
above  the  horizontal.  What  is  the  length  of  a  beam  placed  horizon- 
tally across  the  arch  6  ft.  from  the  top  ? 

7.  The  cable  of  a  suspension  bridge  hangs  in  the  form  of  a 
parabola.  The  roadway,  which  is  horizontal  and  400  ft.  long,  is 
supported  by  vertical  wires  attached  to  the  cable,  the  longest  wire 
being  80  ft.  and  the  shortest  being  20  ft.  Find  the  length  of  a 
supporting  wire  attached  to  the  roadway  75  ft.  from  the  middle. 

8.  Any  section  of  a  given  parabolic  mirror  made  by  a  plane 
passing  through  the  axis  of  the  mirror  is  a  parabolic  segment  of 
which  the  altitude  is  6  in.  and  the  length  of  the  base  10  in.  Find 
the  circumference  of  the  section  of  the  mirror  made  by  a  plane 
perpendicular  to  its  axis  and  4  in.  from  its  vertex 


ELLIPSE 


85 


9.  Find  the  equation  of  the  parabola  having  the  line  ic  =  3  as  its 
directrix  and  having  its  focus  at  the  origin  of  coordinates. 

10.  Find  the  equation  of  the  parabola  having  the  line  y  =  —  2  as 
its  directrix  and  having  its  focus  at  the  point  (2,  4). 

31.  Ellipse.  The  locus  of  a  point  the  sum  of  whose  distances 
from  two  fixed  points  is  constant  is  called  an  ellipse.  The  two 
fixed  points  are  called  i\\Q  foci. 

Let  F  and  F'  (Fig.  38)  be  the  two  foci,  and  let  the  distance 
F'F  be  2  c.  Let  the  straight  line  determined  by  F^  and  F  be 
taken  as  the  axis  of  x^  and  the  y 

middle  point  of  F'F  be  taken 
as  0,  the  origin  of  coordinates, 
and  draw  the  axis  OF.  Then 
the  coordinates  of  F'  and  F 


are  respectively  (—  c,  0)  and     a'\    f^ 
(c,0). 

Let  P(x^  y)  be  any  point 
of  the  ellipse,  and  2  a  repre- 
sent the  constant  sum  of  its 
distances  from  the  foci.   Then, 

from  the  definition  of  the  ellipse,  the  sum  of  the  distances  F'F 
and  FF  is  2  a,  and  from  the  triangle  F'FF  it  is  evident  that 
2  a  >  2  (? ;  whence  a>  c. 


By  §  27, 
and 


F'F=^{x^cy^y' 


FP=^(x-cy+y'', 
whence,  from  the  definition  of  the  ellipse, 


^(x  -h  cy+  y''+^(x  -  cy+f=  2  a. 
Clearing  (1)  of  radicals,  we  have 

Dividing  (2)  by  a^—  c^c^^  we  have 


2^.2 


a^e 


(1) 

(2) 


^.  =  1. 


(3) 


86  ALGEBEAIC  FUNCTIONS 

But  since  a>c,  a^—c^  is  a  positive  quantity  which  may  be 
denoted  by  i^  and  (3)  becomes 

Conversely,  if  the  coordinates  of  any  point  P  satisfy  (4),  it 
can  be  shown  that  the  sum  of  the  distances  F'P  and  FP  is  2  a, 
and  hence  P  is  a  point  of  the  ellipse. 

Solving  (4)  for  y  in  terms  of  x,  we  have 

h 


y  =  ±-\(f—x^,  (5) 

a 

It  is  evident  that  the  only  values  which  can  be  assigned  to  x 
must  be  numerically  less  than  a ;  for  if  any  numerically  larger 
values  are  assigned  to  x^  the  corresponding  values  of  y  are 
imaginary,  and  no  corresponding  points  can  be  plotted.  Hence 
the  curve  lies  entirely  between  the  lines  x  =  —  a  and  x  =  a. 

We  may,  then,  assign  the  possible  values  to  rr,  compute  the 
corresponding  values  of  y,  and,  locating  the  corresponding  points, 
draw  a  smooth  curve  through  them.  As  in  the  case  of  the  pa- 
rabola, we  observe  that  OX  is  an  axis  of  symmetry  of  the  ellipse. 

We  may  also  solve  (4)  for  x  in  terms  of  ?/,  with  the  result 

a:  =  ±^V6^i:7.  (6) 

From  this  form  of  the  equation  we  find  that  the  ellipse  lies 
entirely  between  the  lines  y  =—h  and  y  =  h  and  is  symmetrical 
with  respect  to  OY. 

Hence  the  ellipse  has  two  axes.  A' A  and  B'B  (Fig.  38),  which 
are  at  right  angles  to  each  other.  But  A' A  =  2  a  and  B'B  =  1h\ 
and  since  a  >  6,  it  follows  that  A' A  >  B'B.  Hence  A' A  is  called 
the  major  axis  of  the  ellipse,  and  B'B  is  called  the  minor  axis 
of  the  ellipse. 

The  ends  of  the  major  axis,  A'  and  A,  are  called  the  vertices  of 
the  ellipse,  and  the  point  midway  between  the  vertices  is  called 
the  center  of  the  ellipse ;  that  is,  0  is  the  center  of  the  ellipse, 
and  it  can  be  readily  shown  that  any  chord  of  the  ellipse  which 
passes  through  0  is  bisected  by  that  point. 


ELLIPSE  8T 


From  the  definition  oi  h,  c  =  Va^—  h'\  and  the  coordinates  of 

the  foci  are  (±  Va^—  b%  0). 

OF 
The  ratio  — —  (that  is,  the  ratio  of  the  distance  of  the  focus 

from  the  center  to  the  distance  of  either  vertex  from  the  center) 
is  called  the  eccejitricity  of  the  ellipse  and  is  denoted  by  e.    But 


OF=^a'-b'\  (7) 


and  hence  e  = ;  (S^ 

a  ^  ^ 

whence  it  follows  that  the  eccentricity  of  an  ellipse  is  always 
less  than  unity. 

Similarly,  any  equation  in  form  (4),  in  which  h^  >  a^,  represents 
an  ellipse  with  its  center  at  0,  its  major  axis  on  OF,  and  its 
minor  axis  on   OX.    Then  the  vertices  are  the  points  (0,  ±  5), 


the  foci  are  the  points  (O,  ±  V^^—  d\  and  e  = 

0 

Li  either  case  the  nearer  the  foci  approach  coincidence,  the 
smaller  e  becomes  and  the  more  nearly  h  =  a.  Hence  a  circle 
may  he  considered  as  an  ellipse  with  coincident  foci  and  equal  axes. 
Its  eccentricity  is,  of  course,  zero. 

EXERCISES 

Plot  the  following  ellipses,  finding  the  vertices,  the  foci,  and  the 
eccentricity  of  each : 

1.  9a;2  +  16/=144.  3.  Sx'' -\- 4.y'' =  2. 

2.  9x^-\-Atf  =  36.  4.  2x^+Si/  =  l. 

6.  Find  the  equation  of  the  ellipse  which  has  its  foci  at  the  points 
(—  2,  0)  and  (6,  0)  and  which  has  the  sum  of  the  distances  of  any 
point  on  it  from  the  foci  equal  to  10. 

6.  Find  the  equation  of  the  ellipse  having  its  foci  at  the  points 
(0,  0)  and  (0,  5)  and  having  the  length  of  its  major  axis  equal  to  7. 

32.  Hyperbola.  77ie  locus  of  a  point  the  difference  of  whose 
distances  from  two  fixed  points  is  constant  is  called  a  hyperbola. 
The  two  fixed  points  are  called  the  foci. 


88 


ALGEBRAIC  FUNCTIONS 


Let  F  and  F'  (Fig.  39)  be  the  two  foci,  and  let  the  distance 
F'F  be  2  c.  Let  the  straight  line  determined  by  F'  and  F  be 
taken  as  the  axis  of  x^  and  the  middle  point  of  F'F  be  taken 
as  0,  the  origin  of  coordinates,  and  draw  the  axis  OF.  Then 
the  coordinates  of  F'  and 
F  are  respectively  (—  c?,  0) 
and  (c?,  0). 

Let  P  (x^  y)  be  any  point 
of  the  hyperbola  and  2  a 
represent  the  constant  dif- 
ference of  its  distances  from 
the  foci.  Then,  from  the 
definition  of  the  hyperbola, 
the  difference  of  the  dis- 
tances F'F  and  FF  is  2  «, 
and  from  the  triangle  F'PF  it  is  evident  that  2  a  <  2  <?,  f or  the 
difference  of  any  two  sides  of  a  triangle  is  less  than  the  third 
side  ;  whence  a  <  c. 


Fig.  39 


By  §  27, 

and 

whence  either 


F'P  =  ^(x  +  cy+y^ 
FP  =  ^(x-cy-\-y'^ 


or 


V(^  -  cy+  f-  ^{x  +  cf^-  f=  2  a 

9 


^(x  H-  cy+f-  ^(x  -  cy+  f: 


a, 


(1) 

(2) 


according  as  FP  or  F'F  is  the  greater  distance. 

Clearing  either  (1)  or  (2)  of  radicals,  we  obtain  the  same 
result : 


(of—  c^)x^-^  ay^—  0 
Dividing  (3)  by  a*—  a^e^,  we  have 


+ 


^.=1 


(3) 


(4) 


But  since  a<c,  a^—  c^  is  a  negative  (quantity  whiclx  may  be 
denoted  by  —  J^  and  (4)  becomes 


r 


1. 


(5) 


HYPERBOLA  89 

Conversely,  if  the  coordinates  of  any  point  P  satisfy  (5),  it 
can  be  shown  that  the  difference  of  the  distances  F'P  and  FF 
is  2  a,  and  hence  P  is  a  point  of  the  hyperbola. 

Solving  (5)  for  y  in  terms  of  x,  we  have 

y  =  ±^^x'-a\      •  (6) 

In  this  equation  we  may  assume  for  x  only  values  that  are 
numerically  greater  than  a,  as  any  other  values  give  imaginary 
values  for  y.  Hence  there  are  no  points  of  the  hyperbola  be- 
tween the  lines  x  =  —  a  and  x  =  a.  The  hyperbola  is  symmetrical 
with  respect  to  OX. 

As  the  values  assigned  to  x  increase  numerically,  the  corre- 
sponding values  of  y  increase  numerically  and  the  corresponding 
points  of  the  hyperbola  recede  from  the  axis  OX.  We  may,  how- 
ever, write  (6)  in  the  form  

.  =  ±^4l-J  (7) 

whence  it  is  evident  that  as  the  values  of  x  increase  numerically, 

the  radical  -vjl ^  approaches  unity  as  a  limit.    Therefore  the 

values  of  y  approach,  for  any  chosen  value  of  rr,  the  values  of  y 
for  the  straight  lines  y  =  -xovy  = x.   Hence,  by  prolonging 

these  straight  lines  and  the  curve  indefinitely,  we  can  make 
them  come  as  near  together  as  we  please,  but  can  never  make 
them  coincide  exactly. 

Now,  when  a  straight  line  has  such  a  position  with  respect 
to  a  curve  that  as  the  two  are  indefinitely  prolonged  the  dis- 
tance between  them  approaches  zero  as  a  limit,  the  straight  line 
is  called  an  asymptote  of  the  curve.    It  follows  that  the  lines 

y  =  -X  and  y  = x  are  asymptotes  of  the  hyperbola,  and  as 

a  a 

they  are  of  considerable  assistance  in  drawing  the  curve,  we 

have  drawn  them  in  Fig.  39. 

If  we  had  solved  (5)  for  x  in  terms  of  y,  the  result  would 

have  been  a   , ^ 

^  =  ±Vj^+y^  (8) 


90  ALGEBRAIC  FUNCTIONS 

from  which  it  appears  that  all  values  may  be  assigned  to  y^  and 
that  0  F  is  also  an  axis  of  symmetry  of  the  hyperbola. 

The  points  A!  and  A  in  which  one  axis  of  the  hyperbola  inter- 
sects the  hyperbola  are  called  the  vertices,  and  the  portion  of  the 
axis  extending  from  A'  to  A  is  called  the  transverse  axis.  The 
point  midway  between  the  vertices  is  called  the  center-,  that  is, 
0  is  the  center  of  the  hyperbola,  and  it  can  readily  be  shown 
that  any  chord  of  the  hyperbola  which  passes  through  0  is 
bisected  by  that  point.  The  other  axis  of  the  hyperbola,  which 
is  perpendicular  to  the  transverse  axis,  is  called  the  conju- 
gate axis.  This  axis  does  not  intersect  the  curve,  as  is  evident 
from  the  figure,  but  it  is  useful  in  fixing  the  asymptotes  and 
thus  determining  the  shape  of  the  curve  for  large  values  of  x. 

From  the  definition  of  5,  <?  =  Va^4-5^,  and  the  coordinates  of 
the  foci  are  (±Va^+i^,  0).    Therefore 


OF^-^a'+h\  (9) 

If  we    define   the   eccentricity  of  the  hyperbola   as  the  ratio 


OF  

—  ?  we  have  Va^-4-i^ 

OA  e=     ""^     ,  (10) 

a  quantity  which  is  evidently  always  greater  than  unity. 
Similarly,  the  equation     ^        ^ 

1-1=1  (11) 

is  the  equation  of  a  hyperbola,  with  its  center  at  0,  its  trans- 
verse axis  on  OF,  and  its  conjugate  axis  on  OX,  Then  the  ver- 
tices  are  the  points  (0,  ±6),  the  foci  are  the  points  (0,  ±V6^-fa^), 

the  asymptotes  are  the  straight  lines  y  =  ±'-x,  and  e  = • 

If  h  =  a,  in  either  (5)  or  (11),  the  equation  of  the  hyperbola 

assumes  the  form 

x'-y''=^a'     or     y''-x'=:a\  (12) 

and  the  hyperbola  is  called  an  equilateral  hyperbola.  The  equa- 
tions of  the  asymptotes  become  y  =  ±x',  and  as  these  lines 
are  perpendicular  to  each  other,  the  hyperbola  is  also  called  a 
rectangular  hyperbola. 


CURVES 


91 


EXERCISES 

Plot  the  following  hyperbolas,  finding  the  vertices,  the  foci,  the 
asymptotes,  and  the  eccentricity  of  each : 

1.  4x^-92/'=  36.  4,x''-7/=S. 

2.  9x2-42/2=36.  5.  2x^-3?/=!, 

3.  Sy^-2x''=6.  6.  4:f-x^=l. 

7.  Find  the  equation  of  the  hyperbola  having  its  foci  at  the  points 
(0,  0)  and  (4,  0),  and  the  difference  of  the  distances  of  any  point  on 
it  from  the  foci  equal  to  2. 

8.  The  foci  of  a  hyperbola  are  at  the  points  (—  4,  2)  and  (4,  2), 
and  the  difference  of  the  distances  of  any  point  on  it  from  the  foci 
is  4.    Find  the  equation  of  the  hyperbola,  and  plot. 

33.  Other  curves.  In  the  discussion  of  the  parabola,  the  ellipse, 
and  the  hyperbola,  the  axes  of  symmetry  and  the  asymptotes 
were  of  considerable  assistance  in  constructing  the  curves  ;  more- 
over, the  knowledge  that  there  could  be  no  points  of  the  curve 
in  certain  parts  of  the  plane  decreased  the  labor  of  drawing 
the  curves.  We  shall  now  plot  the  loci  of  a  few  equations, 
noting  in  advance  whether  the  curve  is  bounded  in  any  direc- 
tion or  has  any  axes  of  symmetry  or  asymptotes.  In  this  way 
we  shall  be  able  to  anticipate  to  a  con- 
siderable extent  the  form  of  the  curve. 

Ex.1,    (y  +  3)2  =  (a:  -  2)2(a:  +  1). 
Solving  for  y,  we  have 

2/  =  -  3  ±  (x  -  2)  Vx  +  1. 

In  the  first  place,  we  see  that  the  only 
values  that  may  be  assigned  to  x  are  greater 
than  —  1,  and  hence  the  curve  lies  entirely  on 
the  positive  side  of  the  line  x  =  —  1.    Further-  j'jq^  40 

more,  corresponding  to  every  value  of  x,  there 

are  two  values  of  y  which  determine  two  points  at  equal  distances  from 
the  line  y  =—S.  Hence  we  conclude  that  the  line  y  =—  3  is  an  axis  of 
symmetry  of  the  curve. 

Assigning  values  to  x  and  locating  the  points  determined,  we  draw  the 
curve  (Fig.  40). 


92 


ALGEBRAIC  FUNCTIONS 


Ex.  2. 

Solving  for  y^  we  have 


xy  —  4. 

4 
y  =  -. 

X 


It  is  evident,  then,  that  we  may  assign  to  x  any  real  value  except  zero, 
in  which  case  we  should  be  asked  to  divide  4  by  0,  a  process  that  cannot 
be  carried  out.  Consequently,  there  can  be  no  point  of  the  curve  on  the 
line  x  =  0;  that  is,  on  OY.  We  may, 
however,  assume  values  for  x  as  near 
to  zero  as  we  wish,  and  the  nearer  they 
are  to  zero,  the  nearer  the  corresponding 
points  are  to  OY;  but  as  the  points 
come  nearer  to  OY  they  recede  along 
the  curve.  Hence  OF  is  an  asymptote 
of  the  curve. 

If  we  solve  for  x,  we  have 


y 


Fig.  41 


and,  reasoning   as  above,  we   conclude 

that  the  line  y  =  0  (that  is,  the  axis  OX)  is  also  an  asymptote  of  the  curve. 

The  curve  is  drawn  in  Fig.  41.  It  is  a  special  case  of  the  curve 
xy  =  kf  where  ^  is  a  real  constant  which  may  be  either  positive  or  negative, 
and  is,  in  fact,  a  rectangular  hyperbola  referred  to  its  asymptotes  as  axes. 

It  is  customary  to  say  that  when  the  denominator  of  a  fraction  is  zero, 
the  value  of  the  fraction  becomes  infinite.  The  curve  just  constructed 
shows  graphically  what  is  meant  by  such 
an  expression. 

Ex.  3.    xy-\-2x  +  y-l=0. 
Solving  for  y,  we  have 
l-2a; 
1  +  X 
from  which  we  conclude  that  the  line 
a;  =  —  1  is  an  asymptote  of  the  curve. 
Solving  for  x,  we  have 
1-y 

^  =  2T-/ 

from  which  we  conclude  that  the  line  ?/  =  —  2  is  also  an  asymptote  of  the  curve. 

We  accordingly  draw  these  two  asymptotes  (Fig.  42)  and  the  curve 
through  the  points  determined  by  assigning  values  to  either  x  or  y  and 
computing  the  corresponding  values  of  the  other  variable. 

The  curve  is,  in  fact,  a  rectangular  hyperbola,  with  the  lines  x  =—1 
and  2/  =  —  2  as  its  asymptotes. 


Fig.  42 


CURVES 


93 


Solving  for  y,  we  have 


whence  it  is  evident  that  the  curve  is  sym- 
metrical with  respect  to  OX.  The  lines  a:  =  0 
and  X  =  2a,  corresponding  to  the  values  of  x 
which  make  the  numerator  and  the  denomi- 
nator of  the  fraction  under  the  radical  sign 
respectively  zero,  divide  the  plane  into  three 
strips;  and  only  values  between  0  and  2a 
can  be  substituted  for  x,  since  all  other  values 
make  y  imaginary.  It  follows  that  the  curve 
lies  entirely  in  the  strip  bounded  by  the  two 
lines  a:  =  0  and  a:  =  2  a. 

By  the  same  reasoning  that  was  used  in 
Exs.  2  and  3,  it  can  be  shown  that  the  line 
a:  =  2  a  is  an  asymptote  of  the  curve. 

The  curve,  which  is  called  a  cissoid,  is  drawn 
in  Fig.  43. 


EXERCISES 

Plot  the  following  curves  : 

1.  y^  =  x\ 

2.  f  =  X^(X  +  4:). 

3.  f=4.(x-S). 

4.  y^=x^-5x  +  6. 

5.  f  =  x(x^—4:). 

6.  ?/2==iC^— 5»2-|_6£C.  12-2^  = 


Fig.  43 


7.  y'^=4.x*-x\ 

8.  xy^  =  4:  —  X. 

9.  xy  =  —  5. 

10.  Sy-xy  =  12. 

11.  xy— 2x-{-4:y=:0, 
+  1 


34.  Theorems  on  limits.  In  obtaining  more  general  formulas 
for  differentiation,  the  following  theorems  on  limits  will  be 
assumed  without  formal  proof: 

1.  The  limit  of  the  sum  of  a  finite  number  of  variables  is  equal 
to  the  sum  of  the  limits  of  the  variables. 

2.  The  limit  of  the  product  of  a  finite  number  of  variables  is 
equal  to  the  product  of  the  limits  of  the  variables. 


94  ALGEBRAIC  FUNCTIONS 

3.  The  limit  of  a  constant  multiplied  hy  a  variable  is  equal  to 
the  constant  multiplied  hy  the  limit  of  the  variable. 

4.  The  limit  of  the  quotient  of  two  variables  is  equal  to  the 
quotient  of  the  limits  of  the  variables,  provided  the  limit  of  the 
divisor  is  not  zero. 

35,  Theorems  on  derivatives.  In  order  to  extend  the  process 
of  differentiation  to  functions  other  than  polynomials,  we  shall 
need  the  following  theorems : 

1.  The  derivative  of  a  constant  is  zero-. 
This  theorem  was  proved  in  §  8. 

2.  The  derivative  of  a  constant  times  a  function  is  equal  to  the 
constant  times  the  derivative  of  the  function. 

Let  uhe  Q,  function  of  x  which  can  be  differentiated,  let  c  be 
a  constant,  and  place  y  =  cu 

Give  X  an  increment  Ax,  and  let  Au  and  Ay  be  the  corre- 
sponding increments  of  u  and  y.    Then 

Ay  =  c(u-{-  Aw)  —  cu  =  c  Au. 


Hence 

Ay_ 

Ax 

Au 
Ax 

and,  by  theorem  3, 

§34, 

Lim^^  = 

Ax 

c  Lim 

Au 
Ax 

Therefore 

dy 
dx 

du 
dx 

by  the  definition  of  a  derivative. 

Ex.  1.    ?/  =  5(x3  +  3a:2  +  l). 

^  =  5  — (a:3  +  3  a:2  +  1)  =  5  (3  a;2  +  6  a:)  =  15  (a:2  +  2  x). 

3.  The  derivative  of  the  sum  of  a  finite  number  of  functions  is 
equal  to  the  sum  of  the  derivatives  of  the  functions. 

Let  u,  V,  and  w  be  three  functions  of  x  which  can  be  differen- 
tiated,  and  let  ^^^-^v-^w. 


DERIVATIVES 


95 


Give  X  an  increment  Aa:,  and  let  the  corresponding  increments 
of  w,  V,  w,  and  1/  be  Au,  Av,  Aw,  and  Ay.    Then 

Ai/  =  (u  +  Au  +  v-^Av-\-w  +  Aw)  —  (^u-}-v  +  w) 

=  Au-\-  Av  -\-  Aw  ; 


whence 


Ay     Au      Av      Aw 

Ax      Ax      Ax      Ax 


Now  let  Ax  approach  zero.    By  theorem  1,  §  34, 

T  .     Av      T  •     Alt  ,  T  •     Av  ,  ^  .     Aw 
Lim  -—  =  Lim f-  Lim  - — \-  Lmi ; 

Ax  Ax  Ax  Ax 

that  is,  by  the  definition  of  a  derivative, 

dy  _du      dv      div 
dx      dx      dx      dx 

The  proof   is  evidently  applicable  to  any  finite  number  of 
functions. 


Ex.  2.  y  =  x^-Zx^-\-2x'^-  7x. 


dx 


4  a;3  -  9  a:2  +  4  a:  -  7. 


4.  The  derivative  of  the  product  of  a  finite  number  of  functions 
is  equal  to  the  sum  of  the  products  obtained  by  multiplying  the 
derivative  of  each  factor  by  all  the  other  factors. 

Let  u  and  v  be  two  functions  of  x  which  can  be  differentiated, 

and  let 

y  =  uv. 

Give  X  an  increment  Ax,  and  let  the  corresponding  increments 
of  u,  V,  and  y  be  Au^  Av,  and  Ay. 


Then 


and 


Ay  =  (u-{-  All)  (v  +  Av)  —  uv 

=  uAv-\-vAu-^Au'Av 

Ay         Av        Au      Au     . 
-f-  =  u— --\- V -—  +  —-'  Av, 
Ax         Ax        Ax      Ax 


96  ALGEBRAIC  FUNCTIONS 

If,  now,  Ax  approaches  zero,  we  have,  by  §  34, 

Lim  —^  =  u  Lim  - — \-v  Lim  - — h  Lim  — -  •  Lim  Av. 

Ax  Ax  Ax  Ax 

But  Lim  Av  =  0, 

and  therefore  -^  =  u—--{-v—— 

ax         ax        ax 

Again,  let  ?/  =  uvw. 

Regarding  uv  as  one  function  and  applying  the  result  already 
obtained,  we  have 


dy  dw  .      d(uv) 

-^  —  uv-—-\-w    \    ^ 
dx  dx  dx 


dw  , 
uv—--\-w 
dx 


\    dv        duA 
L    dx        dx] 


dw  dv  ,        du 

==  uv  -—  -\-  uw  -—  -\-  vw  —-' 
dx  dx  dx 

The  proof  is  clearly  applicable  to  any  finite  numbers  of  factors. 

Ex.3.    y  =  (Sx-5)(x'^  +  l)x\ 

-^  =  (3  a:  —  0)  (a;2  + 1)     \    ^  +  (3  a:  —  o)  a;^  — ^^— +  (a:^  + 1 )  a:^  -^^ — ; ^ 

dx  dx  dx  dx 

=  (3  X  -  5)  (ar2  +  1)  (3  x^)  +  (3  a:  -  b)x^{2  x)  +  (a:^  +  l)a;8(3) 

=  (18 a:^  -  25  a;2  +  12  a:  -  15)a:2. 

5.  The  derivative  of  a  fraction  is  equal  to  the  denominator  times 
the  derivative  of  the  numerator  minus  the  numerator  times  the  deriva- 
tive of  the  denominator,  all  divided  by  the  square  of  the  denominator. 

Let  y  —  —>  where  u  and  v  are  two  functions  of  x  which  can  be 

V 

differentiated.    Give  x  an  increment  Ax,  and  let  Au,  Av,  and  Ay 
be  the  corresponding  increments  of  u,  v,  and  y.    Then 

.    w  +  Au      u     v  Au  —  uAv 

Ay  = = 

v-\-Av      V         v^-\-vAv 

and 


Au        Av 
V u  — 

Ay        Ax         Ax 


Ax        V  -\-v  Av 


DERIVATIVES  97 

Now  let  A^  approach  zero.    By  §  34, 

J  .     Au         ^  ,     Av 

V  Lim u  Lim  — - 

^  .     Ay  Ax  Ax 

Lim-^= — p — ; 

Ax  t;  +  ?^ -L-im  Av 


whence 


Ex.4.   y  =  ^ 


du         dv 
dy        dx         dx 

dx  ~  v^ 


x^  +  1 


dy  _  (■r2  +  l)(2a:)-(a:^-l)2a:^        4  a: 
dx  (2:2  +  1)2  ~(a:«  +  l)2' 

6.  The  derivative  of  the  nth  power  of  a  function  is  obtained  by 
multiplying  n  times  the  (n—l^th  power  of  the  function  by  the 
derivative  of  the  function. 

Let  y  =  t^",  where  u  is  any  function  of  x  which  can  be  differ- 
entiated and  n  is  a  constant.    We  need  to  distinguish  four  cases : 

Case  I.  When  n  is  a  positive  integer. 

Give  X  an  increment  Ax^  and  let  Au  and  Ay  be  the  corre- 
sponding increments  of  u  and  y.    Then 

Ay  =  (u  +  Auy  —  u""  I 

whence,  by  the  binomial  theorem. 

At/  =  n?/"-^  Au  +  '^^^~^^  u''-\Auy+  ...  -^(^Auy. 

Ay         „_iAm  ,  nCn  —  l^  Au '  ,    ,.    ,„  .Au 

Ax  Aa:  2  Ax  Ax 

Now  let  Ax,  Au,  Ay  approach  zero,  and  apply  theorems  1  and 

2,  §  34.    The  limit  of  —^  is  -^,  the  limit  of  -^  is  -^,  and  the 

Ax       dx  Ax       dx 

limit  of  all  terms  except  the  first  on  the  right-hand  side  of 
the  last  equation  is  zero,  since  each  contains  the  factor  Au. 
Therefore  ^  ^^ 

dx  dx 


98  ALGEBEAIC  FUNCTIONS 

Case  II.  When  n  is  a  positive  rational  fraction. 

Let  n  =  —-,  where  p  and  q  are  positive  integers,  and  place 

y  =  u\ 
By  raising  both  sides  of  this  equation  to  the  ^th  power,  we  have 

Here  we  have  two  functions  of  x  which  are  equal  for  all 
values  of  x. 

Taking  the  derivative  of  both  sides  of  the  last  equation,  we 
have,  by  Case  I,  since  p  and  q  are  positive  integers, 

n   \dy  „   ^du 

Substituting  the  value  of  y  and  dividing,  we  have 

dy  __p   '^-'^du 
dx      q  dx 

Hence,  in  this  case  also, 

dy  „   T  du 

— ^  =  nu       • 

dx  dx 

Case  III.    When  %  is  a  negative  rational  number. 
Let  n=—m,  where  m  is  a  positive  number,  and  place 

„.       1 


y=--  =-.■ 

div") 

dy             dx 

len                              -/  = 

dx          u^-^ 

(by  5) 

mu       — - 
dx 

(by 

Cases  I  and  II) 

=  —  mu         —-• 
dx 

ence,  in  this  case  also. 

dy         n  \du 
dx               dx 

DERIVATIVES  99 

Case  IV.   When  n  is  an  irrational  number. 

The  formula  is  true  in  this  case  also,  but  the  proof  will  not 
be  given. 

It  appears  that  the  theorem  is  true  for  all  real  values  of  n. 
It  may  be  restated  as  a  working-rule  in  the  following  words; 

To  differentiate  a  power  of  any  quantity^  bring  down  the  exponent 
as  a  coefficient^  write  the  quantity  with  an  exponent  one  less,  and 
multiply  by  the  derivative  of  the  quantity. 

Ex.  5.    y  =  (x-3  +  4  x2  -  5  a:  +  7)^. 

^  =  3  (x3  +  4  a:2  -  5  a:  +  7)2  —  (a:^  +  4  a:2  -  5  a:  +  7) 
dx  dx 

=  3  (3  a:2  +  8  a;  -  5)  (x^  +  4  a;2  -  5  a:  +  7)2. 
Ex.  6.    ?/  =  v^  +  i  =  a:!  +  x-^. 


X'' 


-^  =  -X    3— 3a:- 
dx      3 


3^^ 


Ex.  7.    ?/  =  (a:  +  l)Va;2  +  l. 


dx  dx  dx 

=  (a:  +  1)  [I  (x2  +  1)-  i  .  2  a:]  +  (x^  +  1)^ 

=  ^i^  +  (.^  +  l)^ 
(a:2  +  1)2 

2  a:2  +  a:  +  1 


Va;2  +  1 
Ex.8.   y='l1IZ  =  (-I-)\ 


id 


dx~3\x^-\-l)      dxW-hl) 
^  1  /a:^  +  1\H  -  2  x^ 
~3\     X     I    (x^  +  iy 
1  -  2  a:^ 
3a:^(x3  +  l)^ 


100  ALGEBRAIC  FUNCTIONS 

7,  If  y  is  a  function  of  x,  then  x  is  a  function  of  y,  and  the 
derivative  of  x  with  respect  to  y  is  the  reciprocal  of  the  derivative 
of  y  with  respect  to  x. 

Let  Ax  and  Ay  be  corresponding  increments  of  x  and  y.  It  is 
immaterial  whether  Ax  is  assumed  and  Ay  determined,  or  Ay 
is  assumed  and  Ax  determined.    In  either  case 


whence 


that  is, 


Ax 

1 

Ay 

'"Ay 
Ax 

T   irv*                 — 

1 

Ay 

Lim 

Ax 

dx 

1 

Ty~~ 

~'dy 

dx 


^.  If  y  is  a  function  of  u  and  u  is  a  function  of  x,  then  y  is 
a  function  of  x^  and  the  derivative  of  y  with  respect  to  x  is  equal 
to  the  product  of  the  derivative  of  y  with  respect  to  u  and  the 
derivative  of  u  with  respect  to  x. 

An  increment  Ax  determines  an  increment  Au,  and  this  in  turn 
determines  an  increment  Ay.    Then,  evidently. 


whence  Lim  ^^^  =  Lim  ^^  .  Lim  — 

Ax 


that  is, 
Ex.  9.    7/ 


Ay_ 
Ax 

Ay 
Au 

Au 

Ax 

L 

.     Ay 
Ax 

•-  Lira 

Au 

dy_ 
dx 

dy 
du' 

du 
dx' 

=  u^ 

'  +  3 

w  +  l, 

where  u 

_  1 
x^ 

dy  _ 
dx 

.{2u 

+  3)( 

-a- 

2  +  3a:2 
x^ 

4  +  6a:2 


The  same  result  is  obtained  by  substituting  in  the  expression  for  y  the 
vahie  of  u  in  terms  of  x  and  then  differentiating. 


DERIVATIVES  101 

36.  Formulas.    We  may  now  collect  our  formulas  of  differen- 
tiation in  the  following  table : 


%'"■ 

(1) 

d(cu)         du 
dx           dx 

(2) 

d(u-\-v')      du      dv 
dx           dx      dx 

(3) 

d{uv)          dv         du 
dx            dx         dx 

(-t) 

,  /u\         du         dv 
\vl         dx         dx 

(5) 

dx                 v^ 

d(iC)          „  ,du 
dx                   dx 

(6) 

dx       1 

dy      dy 

(7) 

dx 

dy      dy    du 
dx      du    dx 

(8) 

dy 

dy      du 
dx      dx 

(9) 

du 

Formula  (9)  is  a  combination  of  (7)  and  (8). 

The  first  six  formulas  may  be  changed  to  corresponding  for- 
mulas for  differentials  by  multiplying  both  sides  of  each  equation 
hydx.    They  are  ^^^^^^  ^^^^ 

d(^eu)  =  cdu,  (11) 

d(u  +  v)  =  d^w  +  dv,  (12) 

d(uv')  =  udv -\-vdUy  (13) 

,/w\      vdu  —  udv  ,^  .. 

d(ii'')  =  7iu''-^du,  (15) 


102  ALGEBRAIC  FUNCTIONS 

EXERCISES 

Find  -^  in  each  of  the  following  cases : 

1.  7/  =  (3a:  +  l)(2x2+3a;-2). 


^^^+9  *  *^       ^x-1 

'  y      x^  +  x-j-l'  16.  ?/  =  X V9  -  a 


6.  2/  =  "v^aj^ 


1  17.   v/  =  (cc  +  1)  Vx^  -  1. 


■</;^^ 


18.  y 


4       1  V  a^  +  x^ 

7,    y^x'-X---^--  x-1 


X 


19.  y  = 


8.  v  =  (4ic2H-3cc4-l)'.  Vx2_i 
•^       ^   ^  2  X  —  3 

9.  ?/=  Vx^+4x-^-|-l.  20.   ?/  =  -—=. 

10.  2/  -  (^  +  4)J^  21.  2/  =  (^  +  l)^(x^  +  l)i 

11.  2/  =  (a^-x^)^.  ^2_^^8 

.  22.   y  =  -— ^=. 

12.  2/--7i=-  ^^'+9 

./ 23.   y  —    ,  ^. 

13.  7/  =  V(x^  +  lOx^  +  3)1  -^        ^1  +  x« 

37.  Differentiation  of  implicit  functions.  Consider  any  equa- 
tion containing  two  variables  x  and  y.  If  one  of  them,  as  x^  is 
chosen  as  the  independent  variable  and  a  value  is  assigned  to 
it,  the  values  of  y  are  determined.  Hence  the  given  equation 
defines  y  as  a  function  of  x.  If  the  equation  is  solved  for  y  in 
terms  of  x^  y  is  called  an  explicit  function  of  x.  If  the  equation 
is  not  solved  for  y,  y  is  called  an  implicit  function  of  x.  For 
example,  /+ 3:r^+ 42:./ +  4a;  + 2^  +  4  =  0, 

which  may  be  written 

defines  y  as  an  implicit  function  of  x. 

If  the  equation  is  solved  for  y,  the  result 
y=-2x-l±\^x^-2^ 
expresses  y  as  an  explicit  function  of  x. 


IMPLICIT  FUNCTIONS  103 

If  it  is  required  to  find  the  derivative  of  an  implicit  function, 
the  equation  may  be  differentiated  as  given,  the  result  being  an 
equation  which  may  be  solved  algebraically  for  the  derivative. 
This  method  is  illustrated  in  the  following  examples: 

Ex.  1.    a:2  +  r/2  =  5. 

If  X  is  the  independent  variable, 

that  is,  2a:  +  2?/^  =  0, 

dx 

,  dy  X 

whence  -^  = , 

dx  y 

Or  the  derivative  may  be  found  by  taking  the  differential  of  both  sides, 

as  follows :  1/  2  .     i\       j/kn      a 

d{x^  +  y^)  =  d{p)  =  0\ 

that  is,  2  xdx  +  2  ydy  =  0, 

,  dy  X 

whence  -^  = 

dx  y 

It  is  also  possible  first  to  solve  the  given  equation  for  y,  thus  : 


y  =  ±  V5  —  x^ : 
whence  -^  =  ± 


dx  V5  -  x^ 

a  result  evidently  equivalent  to  the  result  previously  found. 

The  method  of  finding  the  second  derivative  of  an  implicit 
function  is  illustrated  in  the  following  example: 


Ex.  2.    Find   ^   if   a:2  +  2/2  =  5. 
dx^ 


We  know  from  Ex.  1  that 
Therefore 


dy  _      X 
dx  y 

dx^ 


dx\yl 


f 
f 

_      y2  +  a:^  _       5 
I 
since  y"^  -^  x'^  =  5,  from  the  given  equation. 


104  ALGEBRAIC  FUNCTIONS 

EXERCISES 

dy 
Find  -J-  from  each  of  the  following  equations : 


1.  x''^-if-?,axy  =  ^.  3.  y"  ■ 


x-\-y 


2.  x^y  +  4  a?y  =  8  a^.  4.    Vy  +  x  +  V^/  —  cc  —  a. 

ct?/  clj  u 

Find  -f^  and  -7^  from  each  of  the  following  equations : 
(tX  ctx 

5.  2x^-^Sf=6.  .9.  J  +  yi^J, 

6.  4x'2-9  2/2=36. 

8.  x^  -\-y^  =  aK  11.  a;2  _|_  ^^/  _^  2/^  ^  «^. 

38.  Tangent  line.    Let  Il(xj^,  y-^  be  a  chosen  point  of  any 

curve,  and  let  (— )  be  the  value  of  -~  when  x  =  x.  and  y  =  y.' 

/ ,  V         \dx/i  dx  ^ 

Then  (  — )  is  the  slope  of  the  curve  at  the  point  Jff,  and  also 

\dx/i 

the  slope  of  the  tangent  line  (§  15)  to  the  curve  at  that  point. 
Accordingly,  the  equation  of  the  tangent  line  at  P^  is  (§15) 


^-^'=(l)/"-"'>- 


Ex.  1.  Find  the  equation  of  the  tangent  Une  to  the  parabola  ?/2  =  3  ^  at 
the  point  (3,  3). 

By  differentiation  we  have 

24^  =  3; 

dx 

whence  -r  —  tt-' 

dx      2y 

Hence,  at  the  point  (3,  3),  the  slope  of  the  tangent  line  is  ^,  and  its 
equation  is  y_3=,i(^_3) 

or  x-2y  +  S  =0. 

The  angle  of  intersection  of  two  curves  is  the  angle  between 
their  respective  tangents  at  the  point  of  intersection.  The 
method  of  finding  the  angle  of  intersection  is  illustrated  in  the 
example  on  the  following  page. 


TANGENT  LINE 


105 


Ex.  2.    Find  the  angle  of  intersection  of  the  circle  x^  +  3^2  _  g  and  of 
the  parabola  a;^  =  2  y. 

The  points  of  intersection  are  Pi(2,  2) 
and  7^2  (~  2»  ^)  (^^S-  44),  and  from  the  sym- 
metry of  the  diagram  it  is  evident  that  the 
angles  of  intersection  at  P^  and  P^  are  the 
same. 

Differentiating  the  equation  of  the  circle, 

we  have  2a:  +  2?/—  =  0,  whence  -^  = ; 

ax  dx  y 

and  differentiating  the  equation  of  the  pa- 
rabola, we  find  -^  —  X. 
dx 

Hence  at  P^  the  slope  of  the  tangent  to  the  circle  is  —  1,  and  the  slope 
of  the  tangent  to  the  parabola  is  2. 

Accordingly,  if  ^  denotes  the  angle  of  intersection,  by  Ex.  11,  p.  35, 

—  1  —  2 
tan  ^ 


Fig.  44 


iS 


1-2 
tan -13. 


EXERCISES 

1.  Find  the  equation  of  the  tangent  line  to  the  curve  a;'—  83/^ 
+  16  2/  -  8  =  0  at  the  point  (2,  2). 

2.  Find  the  equation  of  the  tangent  line  to  the  curve  5  a;*  —  ^iX^y 
=  4  ?/'  at  the  point  (2, 1). 

3.  Find  the  point  at  which  the  tangent  to  the  curve  8?/  =  ic^  at 
(1,  \)  intersects  the  curve  again. 

4.  Find  the  angle  of  intersection  of  the  tangents  to  the  curve 

y^  =  x^  at  the  points  for  which  x  =  1. 

x^      y^ 

5.  Show  that  the  equation  of  the  tangent  to  the  ellipse  ^  +  7^  =  1 

X  X         1/  1/  do 

at  the  point  (x^,  y^'i^  ~T  +  "ri^  =  1- 


6.  Show   that   the   equation  of   the   tangent   to   the   hyperbola 
2  =  1  at  the  point  (x^,  y^)is^-^  =  l. 


x'      ?/ 


a 


7.  Show  that  the  equation  of  the  tangent  to  the  parabola  y^—kx 

k 
at  the  point  {x^,  y^  is  y^y  =  2  (^  +  ^i)- 

Draw  each  pair  of  -the  following  curves  in  one  diagram  and  deter- 
mine the  angles  at  which  they  intersect : 


106 


ALGEBRAIC  FUNCTIONS 


8.  x''-^y''=5,   x^-{-y''-2x-\-4:2j-5  =  0. 

9.  x^  =  3ij,    9y^=Sx. 

10.  y^=4.x,    x^-\-  f=5. 

11.  y  =  2x,   icy  =  18. 

12.  07  —  4?/  —  4=0,   x^—  4:X  —  4:1/  =  0. 

13.  x^-{-i/==25,    x^=Sy-3, 

39.  The  differentials  dx,  dy,  ds.  On  any  given  curve  let  the 
distance  from  some  fixed  initial  point  measured  along  the  curve 
to  any  point  P  be  denoted  by  s,  where  s  is  positive  if  F  lies  in 
one  direction  from  the  initial  point  and  negative  if  P  lies  in  the 
opposite  direction.  The  choice  of  the 
positive  direction  is  purely  arbitrary. 
We  shall  take  as  the  positive  direc- 
tion of  the  tangent  that  which  shows 
the  positive  direction  of  the  curve, 
and  shall  denote  the  angle  between 
the  positive  direction  of  OX  and  the 
positive  direction  of  the  tangent  by  (/>. 

Now  for  a  fixed  curve  and  a  fixed 
initial  point  the  position  of  a  point  P 
is  determined  if  8  is  given.   Hence  x 
and  «/,  the  coordinates  of  P,  are  functions  of  s  which  in  general 
are  continuous  and  may  be  differentiated.  We  shall  now  show  that 


dx 

-—  =  cos 


dy  _ 


ds 


=  sin  <^. 


Let  2LTC  FQ  =  As  (Fig.  45),  where  P  and  Q  are  so  chosen  that 
As  is  positive.    Then  PP  =  Ax  and  RQ  =  Ay,  and 


Ax 
As 


PR         chord  P^         PP 


arcP^        diVcPQ 

chord  P^  „„„ 

= ^  •  cos  EFQ, 

Ay  _     RQ     _  chord  P() 
As      arcP^        arcP^ 
chord  P^ 


chord  P^ 


RQ 


chord  Pg 


arcP^ 


sinPP^. 


MOTION  IN  A  CURVE 


lOT 


We  shall  assume  without  proof  that  the  ratio  of  a  small  chord 
to  its  arc  is  very  nearly  equal  to  unity,  and  that  the  limit  of 

—  =  1  as  the  point  Q  approaches  the  point  P  along  the 

curve.    At  the  same  time  the  limit  of  EPQ  =  <^.    Hence,  taking 
limits,  we  have       ^^ 


di/  _ 


=  cos  6,         -^ 
as 


sm</). 


(1) 


If  the  notation  of  differentials  is  used,  equations  (1)  become 
dx  =  ds  '  cos  <^,  dy  =  d8  '  sin  </> ; 

whence,  by  squaring  and  adding,  we  obtain  the  important  equation 


ds  =  dx  -\-  dy' 


(2) 


dx    E 


This  relation  between  the  differentials  of  x^  y^  and  s  is  often  rep- 
resented by  the  triangle  of  Fig.  46.  This  figure  is  convenient  as  a 
device  for  memorizing  formulas  (1)  and  (2),  but  it  should  be  borne 
in  mind  that  RQ  \s,  not  rigorously  y 

equal  to  dy  (§  20),  nor  is  PQ  rigor- 
ously equal  to  ds.  In  fact,  BQ  =  Ay, 
and  PQ  =  As;  but  if  this  triangle  is 
regarded  as  a  plane  right  triangle, 
we  recall  immediately  the  values  of 
sin  <^,  cos  ^,  and  tan  <f)  which  have 
been  previously  proved.  '' 

40.  Motion  in  a  curve.  When  a 
body  moves  in  a  curve,  the  discus- 
sion of  velocity  and  acceleration  becomes  somewhat  complicated, 
as  the  directions  as  well  as  the  magnitudes  of  these  quantities 
need  to  be  considered.  We  shall  not  discuss  acceleration,  but 
shall  notice  that  the  definition  for  the  magnitude  of  the  velocity, 
or  the  speed,  is  the  same  as  before  (namely, 

ds 
dt 

where  s  is  distance  measured  on  the  curved  path)  and  that  the 
direction  of  the  velocity  is  that  of  the  tangent  to  the  curve. 


Fig.  46 


108 


ALGEBRAIC  FUNCTIOKS 


Moreover,  as  the  body  moves  along  a  curved  path  through  a 
distance  FQ=As  (Fig.  47),  x  changes  by  an  amount  FE=Ax^ 
and  y  changes  by  an  amount 
ItQ=Ay.    We  have  then 


T- .     As      c^s  ,     .^ 

Lmi  —  =—-  =  v  =  velocity 

At      dt  ^ 


of 


the  body  in  its  path, 
dx 


Lim-— 

A^ 


dt 


component 


-X 


Fig.  47 


ds     dt 


ds 

It 


of  velocity  parallel  to  OX, 

Lim  — ^  =  ^  =  v„  =  component 
At      dt        "  ^ 

of  velocity  parallel  to  OY. 

Otherwise  expressed,  v  represents  the  velocity  of  P,  v^  the 
velocity  of  the  projection  of  P  upon  OX,  and  v^  the  velocity 
of  the  projection  of  P  on  OY. 

Now,  by  (8),  §  36,  and  by  §  39, 

_  dx      dx    ds 

^'''"di 

=  V  cos  <^, 

-  dy     dy 

and  V  =  -^  =  -^ 

'     dt      ds 

==  V  sin  <f). 
Squaring  and  adding,  we  have 

Formulas  (1),  (2),  and  (3)  are  of  especial  value  w^hen  a  par- 
ticle moves  in  the  plane  XOY,  and  the  coordinates  x  and  t/  of 
its  position  at  any  time  t  are  each  given  as  a  function  of  t.  The 
path  of  the  moving  particle  may  then  be  determined  as  follows : 

Assign  any  value  to  t  and  locate  the  point  corresponding  to 
the  values  of  x  and  ?/  thus  determined.  This  will  evidently  be 
the  position  of  the  moving  particle  at  that  instant  of  time.  In 
this  way,  by  assigning  successive  values  to  t  we  can  locate 
other  points  through  which  the  particle  is  moving  at  the  corre- 
sponding instants  of  time.  The  locus  of  the  points  thus  deter- 
mined is  a  curve  which  is  evidently  the  path  of  the  particle. 


C) 


(2) 


(3) 


MOTION  IN  A  CUKVE  109 

The  two  equations  accordingly  represent  the  curve  and  are 
called  its  parametric  representation^  the  variable  t  being  called  a 
parameter, "^  By  (9),  §  36,  the  slope  of  the  curve  is  given  by 
the  formula  ^^ 

^  =  — =  !j^.  (4) 

dx      dx      v^ 

dt 

In  case  t  can  be  eliminated  from  the  two  given  equations,  the 
result  is  the  (:r,  ?/)  equation  of  the  curve,  sometimes  called  the 
Cartesian  equation ;  but  such  elimination  is  not  essential,  and 
often  is  not  desirable,  particularly  if  the  velocity  of  the  particle 
in  its  path  is  to  be  determined. 

Ex.  1.    A  particle  moves  in  the  plane  XOY  so  that  at  any  time  <, 

X  =  a  -\-  ht,         y  =  c  +  dt, 

where  a,  h,  c,  and  d  are  any  real  constants.    Determine  its  path  and  its 
velocity  in  its  path. 

The  path  may  be  determined  in  two  ways : 

From  the  given  equations  we  find  dx  =  hdt,  and  dy  =  ddt-^  whence  —  =  -• 

/  K  dx      0 

As  the  slope  of  the  path  is  always  the  same  (that  is,  -\,  the  path  must  be  a 

straight  line  which  passes  through  the  point  (a,  c)  —  the  point  determined 
when  /  =  0. 

Or  we  may  eliminate  t  from  the  given  equations,  with  the  result 

d 


=  7(^-«)» 


the  equation  of  a  straight  line  passing  through  the  point  (a,  c)  with  the 
slope  -. 

0 

To  determine  the  velocity  of  the  particle  in  its  path  we  find,  by  differ- 
entiating the  given  equations, 

dx      ,  dy      J 

"'  =  71  =  ''     "'^tr^' 

whence,  by  (3),  v  =  Vb^  +  d^. 

Hence  the  particle  moves  along  the  straight  line  with  a  constant  velocity. 

*  It  may  be  noted  in  passing  that  the  parameter  in  the  parametric  represen- 
tation of  a  curve  is  not  necessarily  time,  but  may  be  any  third  variable  in  termi^ 
of  which  X  and  y  can  be  expressed. 


110  ALGEBRAIC  FUNCTIONS 

Ex.  2.  If  a  projectile  starts  with  an  initial  velocity  v^  in  an  initial  direc- 
tion which  makes  an  angle  a  with  the  axis  of  x  taken  as  horizontal,  its 
position  at  any  time  t  is  given  by  the  parametric  equations 

X  =  VqI  cos  a,         y  =  v^t  sin  a  —  \  gt^. 
Find  its  velocity  in  its  path. 

We  have  v^  =  —  =  y^  cos  or, 

dy 

Hence  v  =  Vy  2  _  2  gv^t  sin  a  +  g'^fi. 

EXERCISES 

1.  The  coordinates  of  the  position  of  a  moving  particle  at  any 
time  t  are  given  by  the  equations  x  =  2  t,  y  =  f.  Determine  the  path, 
of  the  particle  and  its  speed  in  its  path. 

2.  The  coordinates  of  the  position  of  a  moving  particle  at  any 
time  t  are  given  by  the  equations  x  =  t'^,  'f/  =  t  -\-  1.  Determine  the 
path  of  the  particle  and  its  speed  in  its  path. 

3.  The  coordinates  of  the  position  of  a  moving  particle  at  any 
time  t  are  given  by  the  equations  x  =  2  t,  1/  =  ^  t  —  ^  t^.  Determine 
the  path  of  the  particle  and  its  speed  in  its  path. 

4.  At  what  point  of  its  path  will  the  particle  of  Ex.  3  be  moving 
most  slowly  ? 

5.  The  coordinates  of  the  position  of  a  moving  particle  at  any 
time  t  are  given  by  the  equations  x  =  t^  —  3,  y  =  f  -\-  2.  Determine 
the  path  of  the  particle  and  its  speed  in  its  path. 

6.  The  coordinates  of  the  position  of  a  moving  particle  at  any 
time  t  are  given  by  the  equations  x  =  At^,  y  =  4:(1  —  ty.  Determine 
the  path  of  the  particle  and  its  speed  in  its  path. 

7.  Find  the  highest  point  in  the  path  of  a  projectile. 

8.  Find  the  point  in  its  path  at  which  the  speed  of  a  projectile 
is  a  minimum. 

9.  Find  the  range  (that  is,  the  distance  to  the  point  at  which 
the  projectile  will  fall  on  OX),  the  velocity  at  that  point,  and  the 
angle  at  which  the  projectile  will  meet  OX. 

10.  Show  that  in  general  the  same  range  may  be  produced  by 
two  different  values  of  a,  and  find  the  value  of  a  which  produces 
the  greatest  range. 

11.  Find  the  (x,  y)  equation  of  the  path  of  a  projectile,  and  plot. 


VELOCITIES  AND  RATES 


111 


41.  Related  velocities  and  rates.  Another  problem  of  some- 
what different  type  arises  when  we  know  the  velocity  of  one 
point  in  its  path,  which  may  be  straight  or  curved,  and  wish  to 
find  the  velocity  of  another  point  which  is  in  some  way  con- 
nected with  the  first  but,  in  general,  describes  a  different  path. 
The  method,  in  general,  is  to  form  an  equation  connecting  the 
distances  traveled  by  the  two  points  and  then  to  differentiate 
the  equation  thus  formed  with  respect  to  the  timet.  The  result 
is  an  equation  connecting  the  velocities  of  the  two  points. 

Ex.  1.  A  lamp  is  60  ft.  above  the  ground.  A  stone  is  let  drop  from  a. 
point  on  the  same  level  as  the  lamp  and  20  ft.  away  from  it.  Find  the  speed 
of  the  stone's  shadow  on  the  ground 
at  the  end  of  1  sec,  assuming  that  the 
distance  traversed  by  a  falling  body  in 
the  time  t  is  16  t'^. 

Let  A  C  (Fig.  48)  be  the  surface  of 
the  ground  which  is  assumed  to  be  a 
horizontal  plane,  L  the  position  of  the 
lamp,  0  the  point  from  which  the  stone 
was  dropped,  and  »S^  the  position  of  the 
stone  at  any  time  t.  Then  Q  is  the  posi- 
tion of  the  shadow  of  S  on  the  ground, 
LSQ  being  a  straight  line.  Let  OS  =  x  and  BQ  =  y.  Then  L  0 
and  BS  =  QO  —  x.   In  the  similar  triangles  LOS  and  SBQ, 

X  _  60  —  a:^ 
20"       v      ' 


20,  BO  =  Q0, 


(1) 


whence  y  =  — 20.  (2) 

X  , 

We  know  a:  =  16  t^,  whence  ~  =  32t;  and  wish  to  find  -^ ,  the  velocity  of  Q. 
dt.  dt 

Differentiating  (2)  with  respect  to  t,  we  have 

dy  _      1200    dx 


When  t 


dt 
dx 


dt 


1  sec,  X  =  16,  and  —  =  32 ;  whence,  by  substitution,  we  find 
dt 

^  =  -  150  ft.  per  second. 

dt  ^ 

The  result  is  negative  because  y  is  decreasing  as  time  goes  on. 
In  §§  6  and  11,  if  the  rate  of  one  of  two  related  quantities 
was  known,  we  were  able  to  find  the  rate  of  the  other  quantity. 


112  ALGEBRAIC  FUNCTIONS 

This  type  of  problem  may  also  be  solved  by  the  same  method 
by  which  the  problem  of  related  velocities  has  been  solved. 
We  shall  illustrate  by  taking  the  same  problem  that  was  used 
in  §11. 

Ex.  2.  Water  is  being  poured  at  the  rate  of  100  cu.  in.  per  second  into 
a  vessel  in  the  shape  of  a  right  circular  cone  of  radius  3  in.  and  altitude 
9  in.  Required  the  rate  at  which  the  depth  of  the  water  is  increasing  when 
the  depth  is  6  in. 

As  in  §  11,  we  have  V  =  ^\  ttJi^  ; 

dV      ,     ,„dh 
whence  -7-  =  i  '"'"■  -r  • 

dt       ^        dt 

dV 
We  have  given  —  =  100,  A  =  6  ;  from  which  we  compute 

dt       IT 

EXERCISES 

1.  A  point  is  moving  on  the  curve  y^  =  x^.  The  velocity  along 
OX  is  2  ft.  per  second.    What  is  the  velocity  along  OY  when  x  =  2? 

2.  A  ball  is  swung  in  a  circle  at  the  end  of  a  cord  3  ft.  long  so 
as  to  make  40  revolutions  per  minute.  If  the  cord  breaks,  allowing 
the  ball  to  fly  off  at  a  tangent,  at  what  rate  will  it  be  receding  from 
the  center  of  its  previous  path  2  sec.  after  the  cord  breaks,  if  no 
allowance  is  made  for  the  action  of  any  new  force  ? 

3.  The  inside  of  a  vessel  is  in  the  form  of  an  inverted  regular 
quadrangular  pyramid,  4  ft.  square  at  the  top  and  2  ft.  deep.  The 
vessel  is  originally  filled  with  water  which  leaks  out  at  the  bottom 
at  the  rate  of  10  cu.  in.  per  minute.  How  fast  is  the  level  of  the 
water  falling  when  the  water  is  10  in.  deep  ? 

4.  The  top  of  a  ladder  20  ft.  long  slides  down  the  side  of  a  ver- 
tical wall  at  a  speed  of  3  ft.  per  second.  The  foot  of  the  ladder  slides 
on  horizontal  land.  Eind  the  path  described  by  the  middle  point  of 
the  ladder,  and  its  speed  in  its  path. 

5.  A  boat  with  the  anchor  fast  on  the  bottom  at  a  depth  of  40  ft. 
is  drifting  at  the  rate  of  3  mi.  per  hour,  the  cable  attached  to  the 
anchor  slipping  over  the  end  of  the  boat.  At  what  rate  is  the  cable 
leaving  the  boat  when  50  ft.  of  cable  are  out,  assuming  it  forms  a 
straight  line  from  the  boat  to  the  anchor  ? 


GENERAL  EXERCISES  113 

6.  A  solution  is  being  poured  into  a  conical  filter  at  the  rate  of 
5  cc.  per  second  and  is  running  out  at  the  rate  of  2  cc.  per  second. 
The  radius  of  the  top  of  the  filter  is  8  cm.  and  the  depth  of  the  filter 
is  20  cm.  Find  the  rate  at  which  the  level  of  the  solution  is  rising 
in  the  filter  when  it  is  one  third  of  the  way  to  the  top. 

7.  A  trough  is  in  the  form  of  a  right  prism  with  its  ends  isosceles 
triangles  placed  vertically.  It  is  5  ft.  long,  1  ft.  across  the  top,  and 
8  in.  deep.  It  contains  water  which  leaks  out  at  the  rate  of  1  qt. 
(57|  cu.  in.)  per  minute.  Find  the  rate  at  which  the  level  of  the 
water  is  sinking  in  the  trough  when  the  depth  is  3  in. 

8.  The  angle  between  the  straight  lines  AB  and  BC  is  60°,  and 
AB  is  40  ft.  long.  A  particle  at  A  begins  to  move  along  AB  toward 
B  at  the  rate  of  5  ft.  per  second,  and  at  the  same  time  a  particle  at 
B  begins  to  move  along  BC  toward  C  at  the  rate  of  4  ft.  per  second.  At 
what  rate  are  the  two  particles  approaching  each  other  at  the  end 
of  1  sec.  ? 

9.  The  foot  of  a  ladder  50  ft.  long  rests  on  horizontal  ground,  and 
the  top  of  the  ladder  rests  against  the  side  of  a  pyramid  which  makes 
an  angle  of  120°  with  the  ground.  If  the  foot  of  the  ladder  is  drawn 
directly  away  from  the  base  of  the  pyramid  at  the  uniform  rate 
of  2  ft.  per  second,  how  fast  will  the  top  of  the  ladder  slide  down 
the  side  of  the  pyramid  ? 

GENERAL  EXERCISES 

Plot  the  curves : 

1.  3x^-^7  f=  21.  9.  2/'(4  +  ic')  =  x\4:-  x"). 

^  10.  xt  =  x^ . 

3.  9a:2- 2/23=16.  «  +  a^ 

4.  if-2y^x''^2x^-l.  11-  y\x^  ^  o?)  =  a?x\ 
8tt^  12.  x^  +  7/=a^. 


5.  ?/  = 


^+^«'  13.  x^^yi  =  a\ 


6.  ay -{- b^x^  =  a%^x\  1 

7.  (^i^-xy=9-x\  ^^'  (2/  +  2)2=^-3pj. 

8.  (x  +  yy=y^(y-^  2).  15.  xY  +  36  =  16  f. 

Find  the  turning-points  of  the  following  curves  and  plot  the  curves : 

16.  y  =  (2  +  x)(4:-  xf.  (^  - 1)'          ,^                 4 

^      ^  ^^  ^         18.  y  =  ^ -^.        19.  y  =  —^ 7. 

17.  y  =  (x  +  3)\x  -  2).  *^         ic  + 1                   -^       x^  -  4 


:ion  IS 


114  ALGEBRAIC  FUNCTIONS 

20.  Find  the  equation  of  the  tangent  to  the  curve  if  =  x^ at 

(O  ft        \  ~l~ 

21.  Find  the  equation  of  the  tangent  to  the  curve  x^  -\-  y^  =  a^ 
at  the  point  (x^,  t/J. 

22.  Prove  that  if  a  tangent  to  a  parabola  'if  =  kx  has  the  slope  m, 

i  k        k  \ 
its  point  of  contact  is   (- — 5,  7; — )   and  therefore  its  equati 
h  \4m^    2  m/  ^ 

tf  =  mx  +  -, —  • 
•^  4  m 

x^       ip- 

23.  Prove  that  if  a  tangent  to  an  ellipse  —  -|-  —  =  1  has  the  slope  m, 

its  point  of  contact  is  (  ±  — ,  -  j  =F     .  r=  j  and  therefore 

its  equation  is  ?/  =  mx  ±  'y/ah'nP  -\-  b". 

24.  Show  that  a  tangent  to  a  parabola  makes  equal  angles  with 
the  axis  and  a  line  from  the  focus  to  the  point  of  contact. 

25.  Show  that  a  tangent  to  an  ellipse  makes  equal  angles  with  the 
two  lines  drawn  to  the  foci  from  the  point  of  contact. 

Find  the  angles  of  intersection  of  the  following  pairs  of  curves : 

26.  y  ~x,   ir  = 


27.  .M-y-20,  y  =  ^^^g 

28.  x^=^y,2x''-^2if=hx. 

29.  x^-^y-^  =  ^,  ic^^  12  2/ -36  =  0. 

30.  2/'  =  ^',  y^  =  (?-^f- 

31.  f={x-?>)\   2/^=16(:r-3). 

32.  2if=x\  x'-\-y^-4.x  =  0. 

33.  The  coordinates  of  a  moving  particle  are  given  by  the  equa- 
tions x  =  f,  y  =  (1—  fy^.  Find  its  path  and  its  velocity  in  its 
path. 

34.  A  particle  moves  so  that  its  coordinates  at  the  time  t  are 

2 
x  =  2t,  y  =  ^        •    Find  its  path  and  its  velocity  in  its  path. 

35.  A  projectile  so  moves  that  x  —  at^  y  =  ht  —  \gf.  Find  its 
path  and  its  velocity  in  its  path. 


GENERAL  EXERCISES  115 

36.  A  body  so  moves  that  x  =  —  2  -^  t\  ij  =  1  -^  t.  Find  its  path 
and  its  velocity  in  its  path. 

37.  A  particle  is  moving  along  the  curve  1/^==  Ax;  and  when  cc  =  4, 
its  ordinate  is  increasing  at  the  rate  of  10  ft.  per  second.  At  what 
rate  is  the  abscissa  then  changing,  and  how  fast  is  the  particle  moving 
in  the  curve  ?  Where  will  the  abscissa  be  changing  ten  times  as  fast 
as  the  ordinate  ? 

38.  A  particle  describes  the  circle  x^  -\-  y^  =  a^  with  a  constant 
velocity  v^.    Find  the  components  of  its  velocity. 

39.  A  particle  describes  the  parabola  if  =  A^ax  in  such  a  way  that 
its  ic-component  of  velocity  is  equal  to  ct.  Find  its  ?/-component  of 
velocity  and  its  velocity  in  its  path. 

40.  A  particle  moves  so  that  x  =  2t,  y  =  2  V^  —  f.  Show  that  it 
moves  around  a  semicircle  in  the  time  from  ^  =  0  to  ^  =  1,  and  find 
its  velocity  in  its  path  during  that  time. 

41.  At  12  o'clock  a  vessel  is  sailing  due  north  at  the  uniform  rate 
of  20  mi.  an  hour.  Another  vessel,  40  mi.  north  of  the  first,  is  sailing 
at  the  uniform  rate  of  15  mi.  an  hour  on  a  course  30°  north  of  east. 
At  what  rate  is  the  distance  between  the  two  vessels  diminishing  at 
the  end  of  one  hour  ?  What  is  the  shortest  distance  between  the 
two  vessels  ? 

42.  The  top  of  a  ladder  32  ft.  long  rests  against  a  vertical  wall, 
and  the  foot  is  drawn  along  a  horizontal  plane  at  the  rate  of  4  ft. 
per  second  in  a  straight  line  from  the  wall.  Find  the  path  of  a 
point  on  the  ladder  one  third  of  the  distance  from  the  foot  of  the 
ladder,  and  its  velocity  in  its  path. 

43.  A  man  standing  on  a  wharf  20  ft.  above  the  water  pulls  in  a 
rope,  attached  to  a  boat,  at  the  uniform  rate  of  3  ft.  per  second.  Find 
the  velocity  with  which  the  boat  approaches  the  wharf. 

44.  The  volume  and  the  radius  of  a  cylindrical  boiler  are  expand- 
ing at  the  rate  of  .8  cu.  ft.  and  .002  ft.  per  minute  respectively.  How 
fast  is  the  length  of  the  boiler  changing  when  the  boiler  contains 
40  cu.  ft.  and  has  a  radius  of  2  f t.  ? 

45.  The  inside  of  a  cistern  is  in  the  form  of  a  frustum  of  a  regular 
quadrangular  pyramid.  The  bottom  is  40  ft.  square,  the  top  is  60  ft. 
square,  and  the  depth  is  10  ft.  If  the  water  leaks  out  at  the  bottom 
at  the  rate  of  5  cu.  ft.  per  minute,  how  fast  is  the  level  of  the  water 
falling  when  the  water  is  5  ft.  deep  in  the  cistern  ? 


116  ALGEBRAIC  FUNCTIONS 

46.  The  inside  of  a  cistern  is  in  the  form  of  a  frustum  of  a  right 
circular  cone  of  vertical  angle  90°.  The  cistern  is  smallest  at  the 
base,  which  is  4  ft.  in  diameter.  Water  is  being  poured  in  at  the  rate 
of  5  cu.  ft.  per  minute.  How  fast  is  the  water  rising  in  the  cistern 
when  it  is  2^  ft.  deep  ? 

47.  The  inside  of  a  bowl  is  in  the  form  of  a  hemispherical  sur- 
face of  radius  10  in.  If  water  is  running  out  of  it  at  the  rate  of 
2  cu.  in.  per  minute,  how  fast  is  the  depth  of  the  water  decreasing 
when  the  water  is  3  in.  deep  ? 

48.  How  fast  is  the  surface  of  the  bowl  in  Ex.  47  being  exposed  ? 

49.  The  inside  of  a  bowl  4  in.  deep  and  8  in.  across  the  top  is  in 
the  form  of  a  surface  of  revolution  formed  by  revolving  a  parabolic 
segment  about  its  axis.  Water  is  running  into  the  bowl  at  the  rate 
of  1  cu.  in.  per  second.  How  fast  is  the  water  rising  in  the  bowl 
when  it  is  2  in.  deep  ? 

50.  It  is  required  to  fence  off  a  rectangular  piece  of  ground  to  con- 
tain 200  sq.  ft.,  one  side  to  be  bounded  by  a  wall  already  constructed. 
Find  the  dimensions  which  will  require  the  least  amount  of  fencing. 

51.  The  hypotenuse  of  a  right  triangle  is  given.  Find  the  other 
sides  if  the  area  is  a  maximum. 

52.  The  stiffness  of  a  rectangular  beam  varies  as  the  product  of  the 
breadth  and  the  cube  of  the  depth.  Find  the  dimensions  of  the  stiffest 
beam  which  can  be  cut  from  a  circular  cylindrical  log  of  diameter  18  in. 

53.  A  rectangular  plot  of  land  to  contain  384  sq.  ft.  is  to  be  in- 
closed by  a  fence,  and  is  to  be  divided  into  two  equal  lots  by  a  fence 
parallel  to  one  of  the  sides.  What  must  be  the  dimensions  of  the 
rectangle  that  the  least  amount  of  fencing  may  be  required  ? 

54.  An  open  tank  with  a  square  base  and  vertical  sides  is  to  have 
a  capacity  of  500  cu.  ft.  Find  the  dimensions  so  that  the  cost  of 
lining  it  may  be  a  minimum. 

55.  A  rectangular  box  with  a  square  base  and  open  at  the  top  is 
to  be  made  out  of  a  given  amount  of  material.  If  no  allowance  is 
made  for  thickness  of  material  or  for  waste  in  construction,  what  are 
the  dimensions  of  the  largest  box  which  can  be  made  ? 

56.  A  metal  vessel,  open  at  the  top,  is  to  be  cast  in  the  form  of  a 
right  circular  cylinder.  If  it  is  to  hold  27  tt  cu.  in.,  and  the  thickness 
of  the  side  and  that  of  the  bottom  are  each  to  be  1  in.,  what  will  be  the 
inside  dimensions  when  the  least  amount  of  material  is  used  ? 


GENERAL  EXERCISES  117 

57.  A  gallon  oil  can  (231  cu.  in.)  is  to  be  made  in  the  form  of  a 
right  circular  cylinder.  The  material  used  for  the  top  and  the  bottom 
costs  twice  as  much  per  square  inch  as  the  material  used  for  the 
side.  What  is  the  radius  of  the  most  economical  can  that  can  be 
made  if  no  allowance  is  made  for  thickness  of  material  or  waste  in 
construction  ? 

58.  A  tent  is  to  be  constructed  in  the  form  of  a  regular  quadran- 
gular pyramid.  Eind  the  ratio  of  its  height  to  a  side  of  its  base  when 
the  air  space  inside  the  tent  is  as  great  as  possible  for  a  given  wall 
surface. 

69.  It  is  required  to  construct  from  two  equal  circular  plates  of 
radius  a  a  buoy  composed  of  two  equal  cones  having  a  common  base. 
Find  the  radius  of  the  base  when  the  volume  is  the  greatest. 

60.  Two  towns,  A  and  B,  are  situated  respectively  12  mi.  and 
18  mi.  back  from  a  straight  river  from  which  they  are  to  get  their 
water  supply  by  means  of  the  same  pumping-station.  At  what  point 
on  the  bank  of  the  river  should  the  station  be  placed  so  that  the  least 
amount  of  piping  may  be  required,  if  the  nearest  points  on  the  river 
from  A  and  B  respectively  are  20  mi.  apart  and  if  the  piping  goes 
directly  from  the  pumping-station  to  each  of  the  towns  ? 

61.  A  man  on  one  side  of  a  river,  the  banks  of  which  are  assumed 
to  be  parallel  straight  lines  \  mi.  apart,  wishes  to  reach  a  point  on 
the  opposite  side  of  the  river  and  5  mi.  further  along  the  bank.  If 
he  can  row  3  mi.  an  hour  and  travel  on  land  5  mi.  an  hour,  find  the 
route  he  should  take  to  make  the  trip  in  the  least  time. 

62.  A  power  house  stands  upon  one  side  of  a  river  of  width  h  miles, 
and  a  manufacturing  plant  stands  upon  the  opposite  side,  a  miles 
downstream.  Find  the  most  economical  way  to  construct  the  con- 
necting cable  if  it  costs  m  dollars  per  mile  on  land  and  n  dollars  a 
mile  through  water,  assuming  the  banks  of  the  river  to  be  parallel 
straight  lines. 

63.  A  vessel  A  is  sailing  due  east  at  the  uniform  rate  of  8  mi. 
per  hour  when  she  sights  another  vessel  B  directly  ahead  and  20  mi. 
away.  B  is  sailing  in  a  straight  course  S.  30°  W.  at  the  uniform  rate 
of  6  mi.  per  hour.  When  will  the  two  vessels  be  nearest  to  each  other? 

64.  The  number  of  tons  of  coal  consumed  per  hour  by  a  certain 
ship  is  0.2  -f  0.001  v*,  where  v  is  the  speed  in  miles  per  hour.  Find 
an  expression  for  the  amount  of  coal  consumed  on  a  voyage  of 
1000  mi.  and  the  most  economical  speed  at  which  to  make  the  voyage. 


118  ALGEBRAIC  FUNCTIONS 

65.  The  fuel  consumed  by  a  certain  steamship  in  an  hour  is  pro- 
portional to  the  cube  of  the  velocity  which  would  be  given  to  the 
steamship  in  still  water.  If  it  is  required  to  steam  a  certain  distance 
against  a  current  flowing  a  miles  an  hour,  find  the  most  economical 
speed. 

66.  An  isosceles  triangle  is  inscribed  in  the  ellipse  ~^  +  f^  =  1, 

(a  >  b),  with  its  vertex  in  the  upper  end  of  the  minor  axis  of  the 
ellipse  and  its  base  parallel  to  the  major  axis.  Determine  the  length 
of  the  base  and  the  altitude  of  the  triangle  of  greatest  area  which 
can  be  so  inscribed. 


CHAPTER  V 

TRIGONOMETRIC  FUNCTIONS 

42.  Circular  measure.  The  circular  measure  of  an  angle  is  the 
quotient  of  the  length  of  an  arc  of  a  circle,  with  its  center  at 
the  vertex  of  the  angle  and  included  between  its  sides,  divided 
by  the  radius  of  the  arc.  Thus,  if  6  is  the  angle,  a  the  length 
of  the  arc,  and  r  the  radius,  we  have 

^=-/  (1) 

The  unit  of  angle  in  this  measurement  is  the  radian,  which 
is  the  angle  for  which  a  =  r  in  (1),  and  any  angle  may  be  said 

to  contain  a  certain  number  of  radians.    But  the  quotient  -  in 

r 

formula  (1)  is  an  abstract  number,  and  it  is  also  customary  to 
speak  of  the  angle  6  as  having  the  magnitude  -  without  using 
the  word  radian.  Thus,  we  speak  of  the  angle  1,  the  angle  J, 
the  angle  —  ?  etc. 

In  all  work  involving  calculus,  and  in  most  theoretical  work 
of  any  kind,  all  angles  which  occur  are  understood  to  be  ex- 
pressed in  radians.  In  fact,  many  of  the  calculus  formulas  would 
be  false  unless  the  angles  involved  were  so  expressed.  The 
student  should  carefully  note  this  fact,  although  the  reason  for 
it  is  not  yet  apparent. 

From  this  point  of  view  such  a  trigonometric  equation  as 

y  =  ^mx  (2) 

may  be  considered  as  defining  a  functional  relation  between  two 
quantities  exactly  as  does  the  simpler  equation  y  —  x^.  For 
we  may,  in  (2),  assign  any  arbitrary  value  to  x  and  determine 
the  corresponding  value  of  y.    This  may  be  done  by  a  direct 

119 


120  TRIGONOMETRIC  FUNCTIONS 

computation  (as  will  be  shown  in  Chapter  VII),  or  it  may  be 
done  by  means  of  a  table  of  trigonometric  functions,  in  which 
case  we  must  interpret  the  value  of  x  as  denoting  so  many  radians. 
One  of  the  reasons  for  expressing  an  angle  in  circular  measure 
is  that  it  makes  true  the  formula 

Lim?i^=l,  (8) 

A-»0         H 

where  the  left-hand   member   of   the  equation  is  to  be   read 

"  the  limit  of  — -—  as  h  approaches  zero  a's  B 

a  limit.  ^^     \ 

To  prove  this  theorem  we  proceed  as       ^/^T)  '^\^^\\ 

follows :  ^^  I  ]  ) 

Let  h  be  the  angle  AOB  (Fig.  49),  r  the  ""^-.^         j  // 

radius  of  the  arc  AB  described  from  0  as  ^^-V 

a  center,  a  the  length  of  AB,  p  the  length  ^^^  ^' 

of  the  perpendicular  BC  from  B  to  OA, 

and  t  the  length  of  the  tangent  drawn  from  B  to  meet  OA 
produced  in  D. 

Revolve  the  figure  on  OA  as  an  axis  until  B  takes  the  position 
B',  Then  the  chord  BCB'=2p,  the  arc  BAB' =2  a,  and  the 
tangent  B'D=  the  tangent  BD.    Evidently 

BD+DB'  >  BAB'  >BCB'; 

whence  t  >  a  >  p. 

Dividing  through  by  r,  we  have 


r      r       r 

that  is. 

tan  h>  h>  sin  h. 

Dividing  by  sin 

h, 

we 

have 

cos  h      sm  h 

or,  by  inverting, 

,  ^  sin  A  ^  ^ 
cos  h  <  —7—  <  1. 

GRAPHS  121 

Now  as  h  approaches  zero,  cos  A  approaches  1.    Hence  — — > 

h 

which  lies  between  cos  A  and  1,  must  also  approach  1;  that  is, 

T  .     sin  ^     ^ 
Lim— z— =1. 

This  result  may  be  used  to  find  the  limit  of as  h 

approaches  zero  as  a  limit.    For  we  have 

r»   •  2^       .  2^         I   •    h 

2  suT  -     sin^  -      ,  I  sm  - 

1-cosA  2  2     A  2 


h 

2 


h         2\     h 
2  1     2    / 

.    h 

sm- 

Now  as  h  approaches  zero  as  a  limit, approaches  unity, 

by  (3).  Therefore  ^ 

A 
-^.     1  — cosA      ^  ,.. 

Lim =  0.  (4) 

43.  Graphs  of  trigonometric  functions.  We  may  plot  a  trigo- 
nometric function  by  assigning  values  to  x  and  computing,  or 
taking  from  a  table,  the  corresponding  values  of  y.  In  so  doing, 
any  angle  which  may  occur  should  be  expressed  in  circular 
measure,  as  explained  in  the  previous  section.  In  this  connec- 
tion it  is  to  be  remembered  that  tt  is  simply  the  number  3.1416, 
and  that  the  angle  tt  means  an  angle  with  that  number  of  radians 
and  is  therefore  the  angle  whose  degree  measure  is  180°. 

The  manner  of  plotting  can  be  best  explained  by  examples. 

Ex.  1.   y  —  a  sin  hx. 

It  is  convenient  first  to  fix  the  values  of  x  which  make  y  equal  to  zero. 
Now  the  sine  is  zero  when  the  angle  is  0,  tt,  2  tt,  3  tt,  —  tt,  —  2  tt,  or,  in 
general,  k-K,  where  k  is  any  positive  or  negative  integer.  To  make  y  =  0, 
therefore,  we  have  to  place  bx  =  kw;  whence 

27r  TT       -       TT      27r      Stt 

x=-", — ,     - -,     0,     -,     --,     — -,     •••. 

b  b  b         b  b 

The  sine  takes  its  maximum  value  +  1  when  the  angle  has  the  values 

7r.5  7r97r^       ^,    ^  .     .    ^,.  ,  ''■5  7r9ir,„., 

-  >  — -  >  — ,  etc. ;  that  is,  in  this  case,  when  ar  =  -— ,  -— ,  — - ,  etc.  h  or  these 
222  2b    2b    2b 

values  oi  X,  y  =  a. 


122 


TKIGONOMETRIC  FUNCTIONS 


The  sine  takes  its  minimum  value  —1  when  the  angle  is  — ^  — ->  etc. ; 

that  is,  in  this  case,  when  x  —  — »  — >  etc.    For  these  values  oi  x,  y  =—  a. 

These  values  of  x  for  which  the  sine  is  ±  1  lie  halfway  between  the 
values  of  x  for  which  the  sine  is  0. 

Y 


Fig.  50 

These  points  on  the  graph  are  enough  to  determine  its  general  shape. 
Other  values  of  x  may  be  used  to  fix  the  shape  more  exactly.  The  graph 
is  shown  in  Fig.  50,  with  a  =  3  and  b  =  2.  The  curve  may  be  said  to  repre- 
sent a  wave.  The  distance  from  peak  to  peak,  — ,  is  the  wave  length,  and 
the  height  a  above  OX  is  the  amplitude. 

Ex.  2.    y  —  a  cosbx. 

As  in  Ex.  1,  we  fix  first  the  points  for  which  y  =  0.  Now  the  cosine  of 
an  angle  is  zero  when  the  angle  is  ->  -— -j  -— -»  etc.;  that  is,  any  odd 
multiple  of  -  •    We  have,  therefore,  y  =  0  when 


Fig.  51 


Halfway  between  these  points  the  cosine  has  its  maximum  value  +  1 
or  its  minimum  value  —  1  alternately,  and  y  —±a.  The  graph  is  shown 
in  Fig.  51,  with  a  =  3  and  6  =  2. 


GRAPHS 


123 


Ex.3,   y  =  asin(bx  +  c). 

We  have  y  =  0  when  bx  +  c  =  0,  tt,  2  tt,  3  rr,  etc.;  that  is,  when 


Fig.  52 

Halfway  between  these  values  of  x  the  sine  has  its  maximum  value  + 1 
and  its  minimum  value  —  1  alternately,  and  y  =±  a.    The  curve  is  the 

same  as  in  Ex.  1,  but  is  shifted  -  units  to  the  left  (Fig.  52). 

Ex.4.    ?/ =  sinx  +  ^  sin  2x. 

The  graph  is  found  by  adding  the  ordinates  of  the  two  curves  y  =  sin  x 
and  y  =  \  sin  2  x,  as  shown  in  Fig.  53. 

Y 


i^-ysinSx 


^---~rl        o 

/S^x    „.. 

■""'X>/ 

2     --'          \ 

\    y=  sin  X  +  -J  sin  2x 
~'=sinx 


Fig.  53 

EXERCISES 

Plot  the  graphs  of  the  following  equations 


1.  ?/  =  2  sin  3  a?. 

2.  2/  =  3cos^. 

3.  ?/  =  5  sin  (^  —  t)' 

4.  ?/  =  2  cos  (^  +  t) 

5.  2/  =  2  sin  (x  -  2). 


e.  y  =  tan  2  a;. 

7.  y  =  ctn  3  a:;. 
8.7/  =  sec  07. 

d.  y  =  CSC  2  a:. 

10.  ?/  =  vers  0!. 

11.  ?/  =  1  +  sin  2  a;. 

12.  y  =  sin^x  -\-  sin  2 x. 


iU  TKIGONOMETHIC  t^UNCTlOKS 

44.  Differentiation  of  trigonometric  functions.  The  formulas 
for  the  differentiation  of  trigonometric  functions  are  as  follows, 
where  u  represents  any  function  of  x  which  can  be  differentiated : 

d    .  du  ^-, . 

-— sinu  =  co&u—-'>  (1) 

dx  dx 

(2) 
(3) 
(4) 
(5) 

(6) 

These  formulas  are  proved  as  follows: 

1.  Let  2/  =  sin  u^  where  u  is  any  function  of  x  which  may  be 
differentiated.  Give  x  an  increment  Ax  and  let  Au  and  A?/  be 
the  corresponding  increments  of  u  and  ?/.    Then 

At/  =  sin  (u  +  Aw)  —  sin  w 

=  sin  u  cos  Aw  +  cos  u  sin  A?i  —  sin  u 

=  cos  u  sin  Aw  —  (1  —  cos  Aw)  sin  w  ; 

Aw  sin  Aw      1— cosAw   . 

whence  — ^  =  cos  w  — sm  w. 

Aw  Aw  Aw 


d 

dx 

COS  w  = 

du 
—  smw-—» 
dx 

d 
dx 

tan  w  = 

2    du 
sec^w  -— , 
dx 

d 
dx 

ctnw  = 

2    du 

—  CSC  W  —-5 

dx 

d 
dx 

sec  w  = 

du 
sec  w  tan  W-— J 
dx 

d 
dx 

CSC  w  = 

du 

—  CSC  w  ctn  w— - 

dx 

Now  let  Ax  and  therefore  Aw  approach  zero.    By  (3),  §  42, 
di/  _ 


Urn  ^^^^  =  1,  and,  by  (4),  §  42,  Lim  -*•     ^^^^"^  =  0.  Therefore 

Aw  '    J  V  y  5      »  ^^ 


cos  w. 
du 


But  by  (8),  §  36,        ^  =  ^^, 
•^  ^  ^    ^  dx      dudx 

and  therefore  -^  =  cos  w  — — 

dx  dx 


DIFFERENTIATION  125 


2.  To  find  — -  cos  u,  we  write 
ax 

cos  u  =  sin 


ing-«). 

Then  -;-  cos  u  =  —-sm(—  —  u] 

ax  ax       \2        / 


=  —  sin  t*  -—■ 


3.  To  find  — -  tan  w,  we  write 
ax 

sm  w 

tan  u  = 

cos  w 
rr^,  d  ^  c?  sin  w 

1  hen     -—  tan  u  =  - 

ax  ax  cos  u 


d    .  .        d 

cos  ^  —  sin  ?^  —  sin  u  —  cos  u 

't _ £f (by  (5),  §  36) 


(o  .   o   ■-  au 

cos  2^  +  sin^w)— - 
dx 


(by  (1)  and  (2)) 


=  sec  M-— 
dx 

4.  To  find  — -  ctnw,  we  write 
dx 

cosw 

ctn  u  —  — 

sm  w 

m,  d      ^  d    GQSU 

1  hen      — -  ctn  u  =  - — -. 

dx  dx  sin  u 


d  d    . 

sinw  —  cos  w  —  cos  w  —  sm  1^ 

ff ff (by  (5),  §  36) 

-^m'u-GO^'u  du  ^^y  ^^^  ^^^  ^2)) 

sin^ifc  .        dx 
du 
dx 


126  TRIGONOMETRIC  FUNCTIONS 

5.  To  find  — -  sec  w,  we  write 
ax 

sec  u  = =  (cos  u)  \ 

cosw 

Then  —  sect^  =  — (cos  w)"^  — cosw         (by  (6),  §  36) 


sin  2^  du 
GO^^u  dx 

du 

^QGUidillU- — 

dx 


(by  (2)) 


6.  To  find  -—  CSC  u^  we  write 
dx 

CSC  u  =  -: —  =  (sm  U)  \ 
smw 

Then  —  esc  u  =  —  (sin  w)~^  —  sin  u         (by  (6),  §  36) 

(XX  (XX 

—  —  CSC  u  ctn  i^  —  •  (By  (1)) 

(XX 

Ex.  1.    y  —  tan  2  a:  —  tan^a:  =  tan  2  x  —  (tan  a:)^. 

—  =  860^2  a;  —  (2  x)  —  2  (tan  x)  —  tan  x 
dx  dx  dx 

—  2  860^2  a;  —  2  tan  x  sec^a:. 

Ex.  2.    y  =  (2  sec^a:  +  3  sec^a:)  sin  x. 

—  =  sin  a:    Ssec^a; — (seca:)+  6  sec  a:  —  (sec  a:)    +  (2sec*a:  +  Ssec^a:)  —  (sin  a:) 
dx  L  dx  dx  J  dx 

=  sin  a:  (8  sec*  a:  tan  a:  +  6  sec^a:  tan  a;)  +  (2  sec*  a:  +  3  sec^a:)  cos  x 

=  (1  —  cos^a;)  (8  sec^a:  +  6  sec^a:)  +  (2  sec^a;  +  3  sec  x) 

—  8  sec^x  —  3  sec  x. 

EXERCISES 

dti 
Eind  "T-  in  each  of  the  followincr  cases : 
dx 

1.  y  =  3  sin  5x. 

2.  y  =  2tan|. 

3.  y  =  J  sin'2jr. 

4.  y  =  cos^Scc. 


5. 

X       1    .     ^ 

2/  =  ---sm2a:. 

6. 

y  =^\  sin^5  £c  —  ;^  sin^5  x. 

7. 

J>x 

8. 

2/  =  J  csc^3  2c. 

SIMPLE  HAKMONIC  MOTION  127 

2       ^x      ^       X  ■  ,    .    X      ^  X 

9.  2/ = -cos^- —  2cos--  11.  i/  =  4:sm-— 2xco8^' 

2    ,   „x   ,   ^    ,    X  ,  ^  sec  ic  +  tan  x 

10.  2/  =  o  ctn^o  +  2  ctn  -.  12.  ?/  = '-- 

,3        2  2  -^       sec  ic  —  tan  a; 

13.  ?/  =  sin  (2  X  +  1)  cos  (2x  —  1). 

14.  y  =  tan^3cc- 3tan3x  +  9a;. 

15.  ?/ =  sec  2  X  tan  2  cc.  t^ 

16.  y  =  5V(3cos52x- 5cos»2a:). 

17.  sin2a;H-tan3?/  =  0. 

18.  xy  -\-  ctn  xt/  =  0. 

45.  Simple  harmonic  motion.  Let  a  particle  of  mass  m  move 
in  a  straight  line  so  that  its  distance  s  measured  from  a  fixed 
point  in  the  line  is  given  at  any  time  t  by  the  equation 

s  =  c  sin  bt,  (1) 

where  c  and  h  are  constants.    We  have  for  the  velocity  v  and 

the  acceleration  a  i        i^  ^^.^ 

v  =  eo  cos  btj  (2) 

a  =  —  (?5^sin  5^.  (3) 

When  ^  =  0,  8  =  0  and  the  particle  is  at  0  (Fig.  54).  When 
t^-^r-p  s  =  e  and  the  particle  is  at  A,  where  0A=  c. 

When  t  is  between  0  and  — ->  t;  is  positive  and  a  is  negative, 

M  0 

SO  that  the  particle  is  moving  from  0  to  ^  with  decreasing  speed. 

When  t  is  between  — -  and  -rfvis>  ~b  0  J 

Zt  b  0 

negative  and  a  is  negative,  so  that 

the   particle   moves  toward   0  with  increasing   speed.    When 

TT 

^  =  — ,  the  particle  is  at  0. 

As  ^  varies  from  ^  to  -— ^»  the  particle  moves  with  decreasing 
speed  from  0  to  B^  where  OB  =  —  c. 

Finally,  as  t  varies  from  9T  ^o  — »  the  particle  moves  back 
from  ^  to  0  with  increasing  speed. 


128  TRIGONOMETRIC  FUNCTIONS 

The  motion  is  then  repeated,  and  the  particle  oscillates  between 
B  and  J,  the  time  required  for  a  complete  oscillation  being,  as 

2  TT 

we  have  seen,  — —  The  motion  of  the  particle  is  called  simple 
harmonic  motion.   The  quantity  c  is  called  the  amplitude,  and  the 

2  TT 

interval  —r-^  after  which  the  motion  repeats  itself,  is  called 

the  period. 

Since  force  is  proportional  to  the  mass  times  the  acceleration, 
the  force  F  acting  on  the  particle  is  given  by  the  formula 

F—  kma  =  —  hmcV^mi  ht=  —  Jcmh^s. 

This  shows  that  the  force  is  proportional  to  the  distance  s 
from  the  point  0,  The  negative  sign  shows  that  the  force  pro- 
duces acceleration  with  a  sign  opposite  to  that  of  s,  and  there- 
fore slows  up  the  particle  when  it  is  moving  away  from  0  and 
increases  its  speed  when  it  moves  toward  0.  The  force  is  there- 
fore always  directed  toward  0  and  is  an  attracting  force. 

If,  instead  of  equation  (1),  we  write  the  equation 

s  =  c  sin  h(t—  t^),  (4) 

the  change  amounts  simply  to  altering  the  instant  from  which  the 
time  is  measured.  For  the  value  of  s  which  corresponds  to  ^  =  ^^ 
in  (1)  corresponds  to  t=t^-{-t^  in  (4).    Hence  (4)  represents 

2  TT 

simple  harmonic  motion  of  amplitude  c  and  period  -— — 
But  (4)  may  be  written 

s  =  c  cos  5^^  sin  bt  —  c  sin  bt^cos  bt, 
which  is  the  same  as 

s=A  sin  bt-\-B  cos  bt,  (5) 

where  A=^  c  cos  bt^,  B  =  —  c  sin  bt^. 

A  and  B  may  have  any  values  in  (5),  for  if  A  and  B  are  given, 
we  have,  from  the  last  two  equations, 

c=^A^  +  B\  tanJ^  =--, 
°         A 

which  determines  e  and  t^  in  (4). 


SIMPLE  HARMONIC  MOTION  129 

Therefore  equation  (5)  also  represents  simple  harmonic  motion 
with  amplitude  y/ A^-\-B'^  and  period  — — 

In  particular,  if  in  (5)  A—0  and  B  =  c^  we  have 

«  =  c  cos  hU  (6) 

If  in  (4)  we  place  t^  =  —-  —  tl^  it  becomes 

8=:CG0^h(t-t[),  (7) 

which  differs  from  (6)  only  in  the  instant  from  which  the  time 
is  measured. 

EXERCISES 

1.  A  particle  moves  with  constant  velocity  v^  around  a  circle. 
Prove  that  its  projection  on  any  diameter  of  the  circle  describes 
simple  harmonic  motion. 

2.  A  point  moves  with  simple  harmonic  motion  of  period  4  sec. 
and  amplitude  3  ft.    Find  the  equation  of  its  motion. 

3.  Given  the  equation  5  =  5  sin  2  t.  Find  the  time  of  a  complete 
oscillation  and  the  amplitude  of  the  swing. 

4.  Find  at  what  time  and  place  the  speed  is  the  greatest  for  the 
motion  defined  by  the  equation  s  =  c  sin  ht.  Do  the  same  for  the 
acceleration. 

5.  At  what  point  in  a  simple  harmonic  motion  is  the  velocity  zero, 
and  at  what  point  is  the  acceleration  zero  ? 

6.  The  motion  of  a  particle  in  a  straight  line  is  expressed  by  the 
equation  s  =  5  —  2  cos^^.  Express  the  velocity  and  the  acceleration 
in  terms  of  s  and  show  that  the  motion  is  simple  harmonic. 

7.  A  particle  moving  with  a  simple  harmonic  motion  of  amplitude 
5  ft.  has  a  velocity  of  8  ft.  per  second  when  at  a  distance  of  3  ft. 
from  its  mean  position.    Find  its  period. 

8.  A  particle  moving  with  simple  harmonic  motion  has  a  velocity 
of  6  ft.  per  second  when  at  a  distance  of  8  ft.  from  its  mean  position, 
and  a  velocity  of  8  ft.  per  second  when  at  a  distance  of  6  ft.  from  its 
mean  position.    Find  its  amplitude  and  its  period. 

9.  A  point  moves  with  simple  harmonic  motion  given  by  the 
equation  s  =  a  —  ^  sin  ct.    Describe  its  motion. 


130  TRIGONOMETRIC  FUNCTIONS 

46.  Graphs  of  inverse  trigonometric  functions.    The  equation 

ir  =  sin  y  (1) 

defines  a  relation  between  the  quantities  x  and  y  which  may  be 
stated  by  saying  either  that  x  is  the  sine  of  the  angle  y  or  that 
the  angle  y  has  the  sine  x.  When  we  wish  to  use  the  latter  form 
of  expressing  the  relation,  we  write  in  place  of  equation  (1) 
the  equation  y  =  ,iu-^^,  (2) 

where  —  1  is  not  to  be  understood  as  a  negative  exponent  but  as 
part  of  a  new  symbol  sin~^  To  avoid  the  possible  ambiguity 
formula  (2)  is  sometimes  written 

y  =  arc  sin  x. 

Equations  (1)  and  (2)  have  exactly  the  same  meaning,  and 
the  student  should  accustom  himself  to  pass  from  one  to  the 
other  without  difficulty.  In  equation  (1)  y  is  considered  the 
independent  variable,  while  in  (2)  x  is  the  independent  variable. 
Equation  (2)  then  defines  a  function  of  x  which  is  called  the 
anti-sine  of  x  or  the  inverse  sine  of  x.  It  will  add  to  the  clearness 
of  the  student's  thinking,  however,  if  he  will  read  equation  (2) 
as  ''  y  is  the  angle  whose  sine  is  a:." 

Similarly,  if  x  =  cosy,  then  y  =  GOS~'^x;  if  x  =  ts^ny,  then 
y  =  tan~^  x ;  and  so  on  for  the  other  trigonometric  functions.  We 
get  in  this  way  the  whole  class  of  inverse  trigonometric  functions. 

It  is  to  be  noticed  that,  from  equation  (2),  y  is  not  completely 
determined  when  x  is  given,  since  there  is  an  infinite  number 

1  TT 

of  angles  with  the  same  sine.  For  example,  if  a:  =  -»  ^  =  ^' 
CIO  Z  b 

— 1  —-—■,  etc.    This  causes  a  certain  amount  of  ambiguity  in 

using  inverse  trigonometric  functions,  but  the  ambiguity  is  re- 
moved if  the  quadrant  is  known  in  which  the  angle  y  lies.  We 
have  the  same  sort  of  ambiguity  when  we  pass  from  the  equa- 
tion x  =  y^  to  the  equation  y  =  ±  Va:,  for  if  x  is  given,  there 
are  two  values  of  y. 

To  obtain  the  graph  of  the  function  expressed  in  (2)  we 
may  change  (2)  into  the  equivalent  form  (1)  and  proceed  as 


I 


GRAPHS 


131 


m  §  43.  In  this  way  it  is  evident  that  the  graphs  of  the  inverse 
trigonometric  functions  are  the  same  as  those  of  the  direct  func- 
tions but  differently  placed  with   reference  to   the  coordinate 

axes.  It  is  to  be  noticed  particularly 
that  to  any  value  of  x  corresponds  an 
infinite  number  of  values  of  y. 


Fig.  56 

Ex.  1.    ?/  =  sin-^x. 

From  this,  x  =  sin  y,  and  we  may  plot  the  graph  by  assuming  values  of 
y  and  computing  those  of  x  (Fig.  55). 

Ex.  2.    y  =  tan-^a:. 

Then  x  =  tan  y,  and  the  graph  is  as  in  Fig.  56. 


EXERCISES 

Plot  the  graphs  of  the  following  equations : 

l.y  =  t^n-^2x.  3.  1/ =  sm-^(x  —  l).  5.  ?/ =l-f- cos-^rc. 

2.  7/  =  ctn-i3ic.  4.  2/ =  tan-^  (x -f  1).  6.  ij  =  ^tsLU-^x. 

7.  7/  =  cos- ^(a:  — 2).  8.  ij  =  sm-\2x  i-1)  - -- 

47.  Differentiation   of   inverse   trigonometric   functions.     The 

formulas  for  the   differentiation   of   the   inverse  trigonometric 
functions  are  as  follows: 

du 


1.   -— sm  ^u  =  -—= 

dx  Vl- 


,2  dx 


Vl 


when  sin-^tt  is  in  the  first  or  the 
fourth  quadrant; 

—  when  sin~^w  is  in  the  second  or 


^     ^      the  third  quadrant. 


132  TRIGONOMETRIC  FUNCTIONS 

2.  —  cos"^w  = —  when  cos~^w  is  in  the  first  or  the 

-^  ~"  ^  second  quadrant ; 

— -  when  cos"^w  is  in  the  third  or  the 


^  fourth  quadrant. 


g     c?        _j  1      du 

dx  \-\-u^  dx 

^     c?  1    _  \      du 

'  dx  1-^u^  dx 

5.  — -sec"^i^  =  —  —-  when  sec~^w  is  in  the  first  or  the 

dx  n^u^-ldx        third  quadrant; 

-r-  when  sec~^«^  is  in  the  second  or 


uy/u^-1  dx 


the  fourth  quadrant. 


d  .  1         du      ^  1      .     .       1      o 

—  CSC  ^w  = —  when  csc~^w  is  in  the  first  or 

dx  u^u'-ldx    the  third  quadrant; 

— -  when  csc~^M  is  in  the  second  or 


u^u""-!  dx 

The  proofs  of  these  formulas 
1.  If                                        y  = 

the  fourth  quadrant, 
are  as  follows: 
:  sin~^w, 

then 

siny  = 

■u. 

Hence, 
whence 

by  § 

44,        cos  y-^  = 
dx 

dx 

du 
"dx' 

1     du 
cos  y  dx 

But  cos  ?/  =  Vl  —  \^  when  y  is  in  the  first  or  the  fourth  quad- 
rant, and  cos  ?/  =  —  Vl  —  i^  when  y  is  in  the  second  or  the  third 
quadrant. 


2.  If 

y  =  cos~^u, 

en 

cos  y  =  u. 

Hence 

dy     du 
^  dx      dx 

lence 

dy  _        1 

du 


dx         sin  y  dx 


I 


DIFFERENTIATION  133 


But  sin  y  =  Vl  —  u^  when  «/  is  in  the  first  or  the  second  quad- 
rant, and  sin  1/  =  —  Vl  —  u^  when  y  is  in  the  third  or  the  fourth 
quadrant. 

3.  If  1/  =  tan~^2^, 

then  tan  y  =  w. 

Hence  sec^  v  -^  =  — - ; 

ax      ax 


whence 


dy  _      1      du 

dx      1+u^  dx 


4.  If  y  =  ctn~^it, 

then  ctn  y  —  u. 

Hence  —  csc^  ?/ -^  = -- ; 


whence 


dy  _         1      du 
dx         l-{-u^  dx 


5.  If  y  =  sec   ^1^, 

then  sec  y  =  u. 

TT  ^         c?y      du 

Hence  sec  ^  tan  v -^  = -7- ; 

dx      dx 


whence 


dy  _         1  du 

dx      sec  y  tan  y  dx 


But  seQy  =  u,  and  tany  =  Vtt^— 1  when  y  is  in  the  first  or 
the  third  quadrant,  and  tan«/  =  — Vm^— 1  when  y  is  in  the 
second  or  the  fourth  quadrant. 

6.  If  y  =  csG~^u, 

then  CSC  y  —  u. 

TT  ^       <^y      ^^^ 

Hence  —  csc  y  ctn  y-^  =  -—; 

dx      dx 

,  (?V  1  (??/ 

whence  -r^  = ;-• 

dx  csc  y  ctn  ^  aa; 


134  TRIGONOMETRIC  FUNCTIONS 


But  csGi/  =  Uy  and  ctn  ?/ =  Vw^  —  1  when  y  is  in  the  first  or 
the  third  quadrant,  and  ctn  y  —  —  ^u^  —  1  when  y  is  in  the 
second  or  the  fourth  quadrant. 

If  the  quadrant  in  which  an  angle  lies  is  not  material  in  a 
problem,  it  will  be  assumed  to  be  in  the  first  quadrant.  This 
applies  particularly  to  formal  exercises  in  differentiation. 


Ex.  1.  2/  =  sin-i  Vl  —  x^,  where  y  is  an  acute  angle. 


dx      Vl-(l-x2)    dx  Vl-a:2 


This  result  may  also  be  obtained  by  placing  sin-^vl 


Ex.  2.  2/  =  sec  -1 V4  x2  +  4  a;  +  2. 

d 


V  4  a:2  4-  4  a:  +  2 
dy ilx ^^^ 


dx      V4  :c2  +  4  a;  +  2  V(4  x-^  +  4  a:  +  2)  - 1 
4x4-2  1 


(4  a;2  +  4  x  +  2)  (2  a;  +  1)      2  a:^  +  2  a:  +  1 
EXERCISES 


Find  -^  in  each  of  the  following  cases : 

1.  y  =  ^i\\-^^x.  ^■\/x'-\-2x 

.  11.  v  =  cos~^- — • 

2.  2/  =  sin~^-. 
y  =  sm    1—^—.  ^  X 


3.  y  = 


4.  3/  =  cos-^^^^~^-  13-  2/  =  tan-W^^^l  +  cos-^^. 

5.  2/  =  tan-i(^  +  l).  14-  2/  =  ^Vrir^^  +  sin-ix. 

6.  y  =  tan- Vi^32^.  15  _^  +  tan-^. 

1  -^      x'2  +  4  2 

7.  2/  =  ctn-i-^-  

^  16.  2/=  Vx2-4-2sec-i-. 

8.  2/  =  sec~^5a!.  ^ 

9.  2/  =  csc-i2a;.  17.  2/  =  tan-i-(^---j. 

•  2/=    an     ^^_  2-  ^g^  y=  Vl— ic^-l-^'cos-i  Vl— ic^. 


I 


ANGULAR  VELOCITY  135 


48.  Angular  velocity.  If  a  line  OP  (Fig.  57)  is  revolving  in 
a  plane  about  one  of  its  ends  0,  and  in  a  time  t  the  line 
OP  has  moved  from  an  initial  position  OM  to  the  position  OP, 
the  angle  MOP  =  6  denotes  the  amount  of  rotation.  The  rate 
of  change  of  6  with  respect  to  ^  is 
called  the  angular  velocity  of  OP.  The 
angular  velocity  is  commonly  denoted 
by  the  Greek  letter  (o ;  so  we  have  y 

the  formula  ^q 

In  accordance  with  §  42  the  angle  6  /- 
is  taken  in  radians ;  so  that  if  t  is  in  ^^^  57 

seconds,   the    angular  velocity  is  in 

radians  per  second.  By  dividing  by  2  tt,  the  angular  velocity 
may  be  reduced  to  revolutions  per  second,  since  one  revolution 
is  equivalent  to  2  7r  radians. 

A  point  Q  on  the  line  OP  at  a  distance  r  from  0  describes 
a  circle  of  radius  r  which  intersects  OM  at  A.  If  s  is  the  length 
of  the  arc  of  the  circle  AQ  measured  from  J,  then,  by  §  42, 

s  =  rO.  (2) 

Now  —  is  called  the  linear  velocity  of  the  point  Q^  since  it 

measures  the  rate  at  which  s  is  described  ;  and  from  (2)  and  (1), 

d%        dO  ^ox 

showing  that  the  farther  the  point  Q  is  from  0  the  greater  is 

its  linear  velocity. 

Similarly,  the  angular  acceleration,  which  is  denoted  by  a,  is 

defined  by  the  relation  ,         ,2/, 

•^  d(o      du  ^.^ 

dh 
This  is   connected   with  the    linear   acceleration  —  by  the 

formula  ^^        ^,^ 


136 


TRIGONOMETRIC  FUNCTIONS 


Ex.  1.  If  a  wheel  revolves  so  that  the  angular  velocity  is  given  by  the 
formula  w  =  8  <,  how  many  revolutions  will  it  make  in  the  time  from  t  =  2 
to  <  =  5  ? 

We  take  a  spoke  of  the  wheel  as  the  line  OP.   Then  we  have 

de  =  8tdt. 

Hence  the  angle  through  which  the  wheel  revolves  in  the  given  time  is 


e=  f%tdt  =  [4  t^t  =  100  -  16  =  84. 

«/2 


The  result  is  in  radians.    It  may  be  reduced  to  revolutions  by  dividing  by 
2  TT.    The  answer  is  13.4  revolutions. 

Ex.  2.  A  particle  traverses  a  circle  at  a  uniform  rate  of  n  revolutions 
a  second.  Determine  the  motion  of  the  projection  of  the  particle  on  a 
diameter  of  the  circle. 

Let  P  (Fig.  58)  be  the  particle, 
OX  the  diameter  of  the  circle,  and 
M  the  projection  of  P  on  OX.  Then 

X  =  a  cos  6, 

where  a  is  the  radius  of  the  circle. 
By  hypothesis  the  angular  velocity 
of  OP  is  2  nir  radians  per  second. 
Therefore 


whence 


dd      ^ 
(0  =  - —  =  2  nir: 
dt 

e  =  2n7rt  +  C. 


Fig.  58 


If  we  consider  that  when  /  =  0,  the  particle  is  on  OX,  then  C  =  0. 

Therefore  ^  „      , 

X  =  a  cos  a  =  a  cos  2  mrt  =  a  cos  o>^ 

The  point  M  therefore  describes  a  simple  harmonic  motion.    In  fact, 
simple  harmonic  motion  is  often  defined  in  this  way. 


EXERCISES 

1.  A  flywheel  4  ft.  in  diameter  makes  3  revolutions  a  second. 
Find  the  components  of  velocity  in  feet  per  second  of  a  point  on  the 
rim  when  it  is  6  in.  above  the  level  of  the  center  of  the  wheel. 

2.  A  point  on  the  rim  of  a  flywheel  of  radius  5  ft.  which  is  3  ft. 
above  the  level  of  the  center  of  the  wheel  has  a  horizontal  component 
of  velocity  of  100  ft.  per  second.  Find  the  number  of  revolutions 
per  second  of  the  wheel. 


I 


CYCLOID 


137 


3.  If  the  horizontal  and  vertical  projections  of  a  point  describe 
simple  harmonic  motions  given  by  the  equations 

x  =  5  cos  St,        y  =  5  sin  3 1, 

show  that  the  point  moves  in  a  circle  and  find  its  angular  velocity. 

49.  The  cycloid.  If  a  wheel  rolls  upon  a  straight  line,  each 
point  of  the  rim  describes  a  curve  called  a  cycloid. 

Let  a  wheel  of  radius  a  roll  upon  the  axis  of  x,  and  let  C 
(Fig.  59)  be  its  center  at  any  time  of  its  motion,  .JV  its  point  of 


Fig.  59 


contact  with  OX,  and  P  the  point  which  describes  the  cycloid. 
Take  as  the  origin  of  coordinates,  O,  the  point  found  by  rolling 
the  wheel  to  the  left  until  F  meets  OX. 


Then 


ON=^tgFN, 


Draw  MP  and  ON,  each  perpendicular  to  OX,  PR  parallel  to 
OX,  and  connect  C  and  P.    Let 

angle  NOP  =  <^. 
Then  x=^OM=ON-  MN 

=^^vcPN-PR 
=  a(t>  —  a  sin  <^. 
i/  =  MP=NC-PC 
=  a  —  a  cos  (j). 
Hence  the  parametric  representation  (§40)  of  the  cycloid  is 
x  =  a(^<j>  —  sm  </)), 
y  =  a(l  —  cos<^). 


138  TRIGONOMETRIC  FUNCTIONS 

If  the  wheel  revolves  with  a  constant  angular  velocity  o)  =  — ' 
we  have,  by  §  40,  ^^^ 

v^.  —  a  (1  —  cos  <!>)-—  =  a(o(\  —  cos  <^), 

v^^  =  a  sni  9  ~  =  aca  sni  9  ; 

whence  v^  =  aV  (2  —  2  cos  <^)  =  4  aV  sin^  ?» 

and  ?;  =  2  «ft)  sin  ^? 

as  an  expression  for  the  velocity  in  its  path  of  a  point  on  the 
rim  of  the  wheel. 

EXERCISES 

1.  Prove  that  the  slope  of  the  cycloid  at  any  point  is  ctn  ^. 

2.  Show  that  the  straight  line  drawn  from  any  point  on  the  rim 
of  a  rolling  wheel  perpendicular  to  the  cycloid  which  that  point  is 
describing  goes  through  the  lowest  point  of  the  rolling  wheel. 

3.  Show  that  any  point  on  the  rim  of  the  wheel  has  a  horizontal 
component  of  velocity  which  is  proportional  to  the  vertical  height  of 
the  point. 

4.  Show  that  the  highest  point  of  the  rolling  wheel  moves  twice 
as  fast  as  either  of  the  two  points  whose  distance  from  the  ground  is 
half  the  radius  of  the  wheel. 

5.  Show  that  the  vertical  component  of  velocity  is  a  maximum 
when  the  point  which  describes  the  cycloid  is  on  the  level  of  the 
center  of  the  rolling  wheel. 

6.  Show  that  a  point  on  the  spoke  of  a  rolling  wheel  at  a  distance 
h  from  the  center  describes  a  curve  given  by  the  equations 

X  =  a<f>  —  h  sin  <f>,  y  =  a  —  h  cos  <^, 

and  find  the  velocity  of  the  point  in  its  path.    The  curve  is  called  a 
frochold. 

7.  Find  the  slope  of  the  trochoid  and  find  the  point  at  which  the 
curve  is  steepest. 

8.  Show  that  when  a  point  on  a  spoke  of  a  wheel  describes  a 
trochoid,  the  average  of  the  velocities  of  the  point  when  in  its  highest 
and  lowest  positions  is  equal  to  the  linear  velocity  of  the  wheel. 


CUIIVATUIIE 


139 


50.  Curvature.  If  a  point  describes  a  curve,  the  change  of 
direction  of  its  motion  may  be  measured  by  the  change  of  the 
angle  </>  (§  15). 

For  example,  in  the  curve  of  Fig.  60,  if  AI^=  s  and  J^I^  =  As, 
and  if  <^i  and  (f)..^  are  the  values  of  </>  for  the  points  j^  and  li 
respectively,  then  </>.^—  c^^  is  the  total  change  of  direction  of  the 
curve  between  J^  and  ^.   If  y 

(^2  —  <^i  =  A(/>,  expressed  in 
circular  measure,  the  ratio 

^  is  the  average  change 

of  direction  per  linear  unit 
of  the  arc  ij^.  Regarding 
(^  as  a  function  of  s,  and 

taking  the  limit  of  —  as 

As 

As   approaches    zero    as    a 

limit,  we  have  -— ?  which  is  called  the  curvature  of  the  curve  at 
as 

the  point  P.   Hence  the  curvature  of  a  curve  is  the  rate  of  change 

of  the  direction  of  the  curve  with  respect  to  the  length  of  the  arc. 


y   X 

yy 

A  — 

^^ 

X^2 

^ 

0 

Fig.  60 

If 


# . 


is  constant,  the  curvature  is  constant  or  uniform ;  other- 
as 

wise  the  curvature  is  variable.    Applying  this  definition  to  the 
circle  of  Fig.  61,  of  which  the 
center  is  C  and  the  radius  is  a, 
we  have  A<^— iJCiJ,  and  hence 

As  =  a  A(f).    Therefore    -~-  —  -  • 

d^      \  As      a 

Hence  -p-  =  -?  and  the  circle  is 

as      a 

a  curve  of  constant  curvature  equal 
to  the  reciprocal  of  its  radius. 

The  reciprocal  of  the  curva- 
ture is  called  the  radius  of  cur- 
vature and  will  be  denoted  by  /?.  Through  every  point  of  a  curve 
we  may  pass  a  circle  with  its  radius  equal  to  /o,  which  shall  have 
the  same  tangent  as  the  curve  at  the  point  and  shall  lie  on  the 


Fig.  61 


140  TRIGONOMETKIC  FUNCTIONS 

same  side  of  the  tangent.  Since  the  curvature  of  a  circle  is 
uniform  and  equal  to  the  reciprocal  of  its  radius,  the  curvatures 
of  the  curve  and  of  the  circle  are  the  same,  and  the  circle  shows 
the  curvature  of  the  curve  in  a  manner  similar  to  that  in  which 
the  tangent  shows  the  direction  of  the  curve.  The  circle  is 
called  the  circle  of  curvature. 

From  the  definition  of  curvature  it  follows  that 

If  the  equation  of  the  curve  is  in  rectangular  coordinates, 

ds 

by  (9),  §36,  p  =  ^. 

dx 
To  transform  this  expression  further,  we  note  that 

ds  =  dx  4-  dy  ; 
whence,  dividing  by  dx    and  taking  the  square  root,  we  have 


dx 


Since  4>  =  tan-  >  {^\  (by  §  15) 

d(^  _       dx^ 

Substituting,  we  have        p  = -r^ • 

In  the  above  expression  for  p  there  is  an  apparent  ambiguity  of 
sign,  on  account  of  the  radical  sign.  If  only  the  numerical  value 
of  p  is  required,  a  negative  sign  may  be  disregarded. 


CURVATURE  141 


and 


Ex.  1.    Find  the  radius  of  curvature  of  the  ellipse H  —  =  1. 

Here  -j.  =  - —- 

ax  a^y 

tPy__    b* 


Therefore  (oV+M^t. 

'^  a*b* 

Ex.  2.    Find  the  radius  of  curvature  of  the  cycloid  (§  49). 

We  have  -—  =  a (1  —  cos  <f>)  =  2a  sin^  — , 

dcp  2 

dy  •      .       o       •     <^         ^ 

-—  =  a  sin  <p  =  2  a  sm  ^  cos  ^• 
dcfi  2         2 

Therefore,  by  (9),  §  36, 

^  =  ctn^. 
dx  2 

Hence  -rl  =  —  ?; csc^^ -^  =  —  - — csc*^, 

dx^  2         2   dx  4rt         2 


and 


(l  +  ctn2<^)5  .    <^ 

=  .2^ ^-^^  =  4  a  sin  ■^' 

1        4<t>  2 


CSC'  ^ 

4a         2 


EXERCISES 

1.  Find  the  radius  of  curvature  of  the  curve  i/  —  -^^  x•^ 

2.  Find  the  radius  of  curvature  of  the  curve  x^  -\-  y^  =  a7'. 

3.  Find  the  radius  of  curvature  of  the  curve  y  =  tan~^(ic  —  1) 
at  the  point  for  which  x  =  2. 

4.  Show  that  the  circle  (x  —  -~\  +  ?/  =  1  is  tangent  to  the  curve 

y  =  sin  X  at  the  point  for  which  ic  =  — ,  and  has  the  same  radius  of 
curvature  at  that  point. 

5.  Find  the  radius  of  curvature  of  the  curve  x  =  cos  t^y  =  cos  2  t^ 
at  the  point  for  which  ^  =  0. 

6.  Find   the    radius    of    curvature    of   the    curve   a?  =  a  cos  <^  + 
a^  sin  <f),  y  =  a  sin  <^  —  a<\>  cos  <j>. 

7.  Show  that  the  product  of  the  radii  of  curvature  of  the  curve 
y  =  ae  ''at  the  two  points  for  which  cc  =  ±  «  is  a^{e  -\-  e"^)^ 


142 


TRIGONOMETRIC  FUNCTIONS 


Pir,e) 


Fig.  62 


51.  Polar  coordinates.  So  far  we  have  determined  the  posi- 
tion of  a  point  in  the  plane  by  two  distances,  x  and  y.  We  may, 
however,  use  a  distance  and  a  direction,  as  follows: 

Let  0  (Fig.  62),  called  the  origin^  or  pole,  be  a  fixed  point,  and 
let  OM,  called  the  initial  line,  be  a  fixed  line.  Take  F  any  point 
in  the  plane,  and  draw  OF.  Denote  OF  by  r,  and  the  angle  MOF 
by  6.  Then  r  and  6  are  called  the  polar  coor- 
dinates of  the  point  F(^r,  6),  and  when  given 
will  completely  determine  F, 

For  example,  the  point  (2,  15°)  is  plotted 
by  laying  off  the  angle  MOF  =15°  and  meas- 
uring 0F=  2. 

OF,  or  r,  is  called  the  radius  vector,  and 
6  the  vectorial  angle,  of  P.  These  quantities  may  be  either 
positive  or  negative.  A  negative  value  of  0  is  laid  off  in  the 
direction  of  the  motion  of  the  hands  of  a  clock,  a  positive  angle 
in  the  opposite  direction.  After  the  angle  6  has  been  constructed, 
positive  values  of  r  are  measured  from  0  along  the  terminal 
line  of  6,  and  negative  values  of  r  from  0  along  the  backward 
extension  of  the  terminal  line.  It  follows  that  the  same  point 
may  have  more  than  one  pair  of  coor- 
dinates. Thus  (2,  195°),  (2,  -165°), 
(-2,  15°),  and  (-2,  -  345°)  refer  to 
the  same  point.  In  practice  it  is  usu- 
ally convenient  to  restrict  6  to  positive 
values. 

Plotting  in  polar  coordinates  is  facili- 
tated by  using  paper  ruled  as  in  Figs.  64 
and  65.  The  angle  0  is  determined  from 
the  numbers  at  the  ends  of  the  straight 

lines,  and  the  value  of  r  is  counted  off  on  the  concentric  circles, 
either  toward  or  away  from  the  number  which  indicates  6, 
according  as  r  is  positive  or  negative. 

The  relation  between  (r,  6)  and  (x,  y)  is  found  as  follows: 

Let  the  pole  0  and  the  initial  line  OM  of  a  system  of  polar 
coordinates  be  at  the  same  time  the  origin  and  the  axis  of  rr  of  a 
system  of  rectangular  coordinates.   Let  F  (Fig.  63)  be  any  point 


Fig.  63 


POLAR  COORDINATES 


143 


of  the  plane,  (x,  y)  its  rectangular  coordinates,  and  (r,  0)  its 

polar  coordinates.    Then,  by  the  definition  of  the  trigonometric 

functions, 

cos  ^  =  -7 
r 


sin 


r 


Whence  follows,  on  the  one  hand, 

x  =  r  cos  ^, 

«/  =  r  sin  ^ ; 
and,  on  the  other  hand, 

sin  e  = ^ 


^x'-\- 


y 


cos  Q  =  — 


(1) 
(2) 


By  means  of  (1)  a  transformation  can  be  made  from  rectangular 
to  polar  coordinates,  and  by  means  of  (2)  from  polar  to  rectangular 
coordinates. 

When  an  equation  is  given  in  polar  coordinates,  the  corre- 
sponding curve  may  be  plotted  by  giving  to  Q  convenient  values, 
computing  the  corre- 
sponding values  of  r,  ^^o  ^^^° 
plotting  the  resulting 
points,  and  drawing  a 
curve  through  them. 

Ex.  1.    r  =  acos^. 

a  is  a  constant  which 
may  be  given  any  con- 
venient value.  We  may 
then  find  from  a  table  of 
natural  cosines  the  value 
of  r  which  corresponds 
to  any  value  of  Q.  By 
plotting  the  points  cor- 
responding to  values  of  0 
from  0°  to  90°,  we  obtain 

the  arc  ABCO  (Fig.  64).  Values  of  B  from  90°  to  180°  give  the  arc  ODEA. 
Values  of  Q  from  180°  to  270°  give  again  the  arc  ABCO,  and  those  from  270° 
to  360°  give  again  the  arc  ODEA .  Values  of  Q  greater  than  360°  can  clearly 
give  no  points  not  already  found.    The  curve  is  a  circle. 


144 


TKIGONOMETRIC  FUNCTIONS 


Ex.  2.    r  =  a  sin  3^. 

As  6  increases  from  0°  to  30°,  r  increases  from  0  to  a ;  as  ^  increases 
from  30°  to  60°,  r  decreases  from  a  to  0 ;  the  point  (r,  6)  traces  out  the  loop 
OAO  (Fig.  65),  which  is  evidently  symmetrical  with  respect  to  the  radius 
OA.  As  6  increases  from 
60°  to  90°,  r  is  negative 
and    decreases    from    0   to 

—  a ;  as  6  increases  from 
90°  to  120°,  r  increases  from 

—  a  to  0 ;  the  point  (r,  6) 
traces  out  the  loop  OBO. 
As  6  increases  from  120° 
to  180°,  the  point  (r,  6) 
traces  out  the  loop  OCO. 
Larger  values  of  6  give 
points  already  found,  since 
sin  3  (180°  +  ^)  = -sin  3  a 
The  three  loops  are  congru- 
ent, because  sin;3  (60°  -\- B)  = 

—  sin  3  0.  This  curve  is  called 
a  rose  of  three  leaves. 


Ex.  3.    r2  =  2  a2  cos  2  0. 
Solving  for  r,  we  have 


±  a  V2  cos  2  e. 


Hence,  corresponding  to  any  values  of  6  which  make  cos  2  6  positive,  there 
will  be  two  values  of  r  numerically  equal  and  opposite  in  sign,  and  two 
corresponding  points  of  the  curve  symmetrically  situated  with  respect 
to  the  pole.  If  values  are  assigned  to  6  which  make  cos  2  6  negative,  the 
corresponding  values  of  r  will  be 
imaginary  and  there  will  be  no 
points  on  the  curve. 

Accordingly,  as  6  increases 
from  0°  to  45°,  r  decreases  numer- 
ically from  aV2  to  0,  and  the 
portions  of  the  curve  in  the  first 
and  the  third  quadrant  are  con- 
structed (Fig.  66)  ;  as  $  increases  from  45°  to  135°,  cos  2  ^  is  negative,  and 
there  is  no  portion  of  the  curve  between  the  lines  6  =  45°  and  0  =  135° ; 
finally,  as  6  increases  from  135°  to  180°,  r  increases  numerically  from  0  to 
a  V2,  and  the  portions  of  the  curve  in  the  second  and  the  fourth  quadrant 
are  constructed.  The  curve  is  now  complete,  as  we  should  only  repeat  the 
curve  already  found  if  we  assigned  further  values  to  ^;  it  is  called  the 
lemniscate. 


GRAPHS 


145 


Fig.  67 


Ex.  4.    The  spiral  of  Archimedes, 

r  =  aO. 

In  plotting,  0  is  usually  considered 
in  circular  measure.  When  6=0,r  =  0', 
and  as  $  increases,  r  increases,  so  that 
the  curve  winds  an  infinite  number  of 
times  around  the  origin  while  reced- 
ing from  it  (Fig.  67).  In  the  figure  the 
heavy  line  represents  the  portion  of 
the  spiral  corresponding  to  positive  values  of  6,  and  the  dotted  line  the 
portion  corresponding  to  negative  values  of  6. 

EXERCISES 

Plot  the  graphs  of  the  following  curves  : 

9.  r  =  a  sin^-- 
o 

10.  7'^=a'^smO. 

11.  7^=  a'^sinSe. 

12.  r  =a(l-cos2^). 

13.  r  =a(l  +  2cos2^). 

14.  r  =  a  tan  0. 

15.  r  =  atsin2e. 
1 


1. 

r  =  a  sin  6. 

2. 

r  =  asm2  0. 

3. 

r  =  a  cos  3  6. 

4. 

.  e 

r  =  a  sin-. 
o 

5. 

e 

r  =  a  cos-- 

6. 

r  =  3cos^  +  5. 

7. 

r  =  3  cos  ^  +  3.* 

8. 

r  =  3  cos  ^  +  2. 

16.  r 


t 


1  H-  cos  ^ 
Find  the  points  of  intersection  of  the  following  pairs  of  curves  : 

17.  r  =2  sin  B,       r  =  2  Vs  cos  Q. 

18.  r^  =  ^2  (jog  Q^      ^.2  ^  ^2  gjj^  2  Q. 

19.  r  =  1  -f  sin  ^,  r  =  2  sin  6. 

20.  r2=ci2sin^,       r'' =  a^  ^\\^?>  Q. 

Transform  the  following  equations  to  polar  coordinates : 

21.  ry  =  4.  23.  x^-\-if—2  ay  =  0. 

22.  x^-\-7/-4.ax-4.ay=  0.  24.  (x^  -f  y^  =  a\x^ -  t/^. 
Transform  the  following  equations  to  rectangular  coordinates : 

25.  r=asec^.  27.  r  =  a  tan  ^. 

26.  r  =  2acos^.  28.  r  =  acos2$. 

*  The  curve  is  called  a  cardioid. 

t  The  curve  is  a  parabola  with  the  origin  at  the  focus. 


146  TEIGONOMETRIC  FUNCTIONS 

52.  The  differentials  dr,  d$y  ds,  in  polar  coordinates.  We  have 
seen,  in  §  39,  that  the  differential  of  arc  in  rectangular  coordinates 
is  given  by  the  equation 

ds'=dx'-^df.  (1) 

If  we  wish  to  change  this  to  polar  coordinates,  we  have  to 

place 

x  =  r  cos  6,         y  =  r  sin  Q  ; 

whence  dx  =  Q.Q'&Odr —  r  mvQdQ^ 

dy  =  sin  6dr  -\-r  cos  Qdd. 

Substituting  in  (1),  we  have 

ds'=dr'^r^dd\  (2) 

This  formula  may  be  remembered  by  means  of  an  "  elemen- 
tary triangle  "  (Fig.  68),  constructed  as  follows : 

Let  P  be  a  point  on  a  curve  r  =/(^),  the  coordinates  of  P 
being  (r,  ^),  where  OP  =  r  and  MOP  =  6.  Let  6  be  increased 
by  an  amount  dO,  thus  determining  another 
point  Q  on  the  curve.  From  (9  as  a  center 
and  with  a  radius  equal  to  r,  describe  an 
arc  of  a  circle  intersecting  OQ  m  R  so  that 
OR  =  OP  =  r.  Then,  by  §  42,  PE  =  rdd.  Now 
BQ  is  equal  to  Ar,  and  P^  is  equal  to  As. 
We  shall  mark  them,  however,  as  dr  and  ds 
respectively,  and  the  formula  (2)  is  then  correctly  obtained  by 
treating  the  triangle  PQE  as  a  right  triangle  with  straight-line 
sides.  The  fact  is  that  the  smaller  the  triangle  becomes  as  Q 
approaches  P,  the  more  nearly  does  it  behave  as  a  straight-line 
triangle ;  and  in  the  limit,  formula  (2)  is  exactly  true. 

Other  formulas  may  be  read  out  of  the  triangle  PQE.  Let  us 
denote  by  ^|r  the  angle  PQE,  which  is  the  angle  made  by  the 
curve  with  any  radius  vector.  Then,  if  we  treat  the  triangle  PQE 
as  a  straight-line  right-angle  triangle,  we  have  the  formulas : 

,       rdO  ,       dr       ,        ,       rdO  ,„. 

sm '»ir  = -— -,     cos'Ur  =  -— ,      tan  lir  =  — —  (6) 

ds  ds  dr 


DIFFERENTIALS  147 

The  above  is  not  a  proof  of  the  formulas.    To  supply  the 
proof  we  need  to  go  through  a  limit  process,  as  follows : 

We  connect  the  points  F  and  ^  by  a  straight  line  (Fig.  69) 
and  draw  a  straight  line  from  P  per- 
pendicular to  0^  meeting   OQ  at  S, 
Then  the  triangle  PQS  is  a  straight- 
line  right-angle  triangle,  and  therefore 

sin  SQP  =   ,    "^f — 
chord  PQ 

SP  arc  PQ 


arc  PQ     chord  PQ 

Fig.  69 
Now  angle  POQ  =  Ad^SiTc  PQ  =  As, 

and,  from  the  right  triangle  OSP,  SP  =  OP  sin  POQ  =  r  sin  A^. 
Therefore 

•     ci^^      rsinA^       arc  PO  sinA^     M       sltg  PQ 

sin  kjCJP  = ■  • —  ^ .  ,  ^    .   (  4  ) 

As         chord  Pg  ^0       As     chord  PQ     ^  ^ 

Now  let  AQ  approach  zero  as  a  limit,  so  that  Q  approaches  P 
along  the  curve.    The  angle  SQP  approaches  the  angle   OPT^ 

where  PT  is  the  tangent  at  P.    At  the  same  time  ^—  ap- 

kQ  tq  Au 

proaches  1,  by   §  42  ;  -—   approaches   —  ->  by  definition ;    and 

arcPO 
-r — ,         approaches  1,  by  §  39.    In  this  figure  we  denote  the 

angle  OPT  by  i|r  and  have,  from  (4), 

sin^  =  r  — ,  (5) 

which  is  the  first  of  formulas  (3).  It  is  true  that  in  Fig.  69  we 
have  denoted  OPT  by  -^  and  that  in  Fig.  68  a/t  denotes  OQP. 
But  if  we  remember  that  the  angle  OQP  approaches  OPT  as  a 
limit  when  Q  approaches  P,  and  that  in  using  Fig.  68  to  read  off 
the  formulas  (3)  we  are  really  anticipating  this  limit  process,  the 
difference  appears  unessential. 

The  other  formulas  (3)  may  be  obtained  by  a  limit  process 
similar  to  the  one  just  used,  or  they  may  be  obtained  more 


148  TRIGONOMETRIC  FUNCTIONS 

quickly  by  combining  (5)  and  (2).    For,  from  (2)  and  (5), 
we  have 


-IMM 


ds  I      \ds 


whence  cos  ^ir  =  -— .  (6) 

ds 

By  dividing  (5)  by  (6)  we  have 

tan  lir  =  — —  (7) 

dr 

In  using  (7)  it  may  be  convenient  to  write  it  in  the  form 

tan  yjr  =  — ?  (8) 

dr 

Td 

since  the  equation  of  the  curve  is  usually  given  in  the  form 

dv 
r  =f(6^^  and  —  is  found  by  direct  differentiation. 
do 

Ex.  Find  the  angle  which  the  curve  r  =  a  sin  4  6  makes  with  the  radius 
vector  6  =  30°. 

Here  — ^  =  4  a  cos  4  ^.  Therefore,  from  (8),  tan  il/  =  _^L^15 =  _  tan  4  0. 

(W  4  a  cos  4  ^      4 

Substituting    6  =  30°,    we    have    tan  i/^  =  |  tan  120°  =  -  |  V3  =  -  .4330. 

Therefore  ij/  =  156°  35'. 

EXERCISES 

1.  Find  the  angle  which  the  curve  r  =  a  cos  3  0  makes  with  the 
radius  vector  6  =  45°. 

2.  Find  the  angle  which  the  curve  r  =  2  -j-  3  cos  0  makes  with  the 
radius  vector  6  =  90°. 

3.  Find  the  angle  which  the  curve  ?' =  a^sin^-  makes  with  the 
initial  line. 

B  0 

4.  Show  that  for  the  curve  r  =  a  sin^  -  ?  xp  =  j.- 

o  o 

5.  Show  that  the  angle  between  the  cardioid  r  =  a(l—  cos  6)  and 
any  radius  vector  is  always  half  the  angle  between  the  radius  vector 
and  the  initial  line. 


GENERAL  EXERCISES  149 

6.  Show  that  the  angle  between  the  leinniscate  r^  =  2  d^  cos  2  6 

TT 

2 


77" 

and  any  radius  vector  is  always  —  plus  twice  the  angle  between  the 


radius  vector  and  the  initial  line. 

7.  Show  that  the  curves  i^  —  ar  sin  2  0  and  i^  =  a^  cos  2  6  inter- 
sect at  right  angles. 

GENERAL  EXERCISES 

Find  the  graphs  of  the  following  equations  : 

,    .    .r  -h  1  .,    .    /         7r\ 

1.  t/  =  4:sm— —  5.  i/  =  Ssm\^x  +  ^j- 

2.  y/  =  cos  (2  a?  —  3).  6.  /  =  tan  x. 

3.  1/  =  tan^.  7.  y  =  2  cos  2(x-  -  2). 

4.  y  =  ^sin  2a; -f- ^sin  3a:.  8.   // =  3  cos 3(x -1- j  ). 

Find  -r^  in  the  following  cases : 
dx 

9.  v/  =  2  a;  +  tan  2  0-. 

10.  y  =  itan(3a;  +  2)-|-itan«(3£c  +  2). 

11.  ij  =  ^x  —  ^^^\\i{^  —  *6 x). 

12.  tan  (x  -\-i/)  +  tan  (x  —  */)  =  0. 

13.  y/  =  3ctn^'^  +  o  ctn^^-  21.  //  =  ctn-^  \i-- 

o  o  yx 

14.  y/ =  csc^4a: -|- 2ctn  4  a;.  22.    y  =  sec~^7a-. 

15.  fy  =  ^tan'^aa;.  ir- -  4 

16.  y/  =  sinHa;cos*2a;.  23.  .y  =  cos-^^qp^- 
2       „a; 


a; 


17.  y  = -cos*7  —  2cos  7-  24.   y  =  ctn"^  Va;^  —  2a;. 
^34               4  ^ 

2 

18.  2/  =  ^tan^2x  — ^tan2a;  +  a:.  25.  y  =  csc-*^ r- 

19.  y  =  sin-^^^-^.  2g^  .y=:3V9^:^^-f  2sin-- 

20.  y  =  cos-^^'^~    ■  27.  y  =  tan'^o;  Va;^  -  2. 

i5 

28.  A  particle  moves  in  a  straight  line  so  that  s  =  5  —  4  sin^— • 

Show  that  the  motion  is  simple  harmonic  and  find  the  center  about 
which  the  particle  oscillates  and  the  amplitude  of  the  motion. 


150  TRIGONOMETRIC  FUNCTIONS 

29.  A  particle  moves  on  the  ellipse  —  -f-  ^-  =  1  so  that  its  projec- 
tion upon  OX  describes  simple  harmonic  motion  given  hy  x  =  a  cos  kt. 
Show  that  its  projection  upon  OY  also  describes  simple  harmonic 
motion  and  find  the  velocity  of  the  particle  in  its  path. 

30.  A  particle  moving  with  simple  harmonic  motion  of  period  ~ 

o 
has  a  velocity  of  9  ft.  per  second  when  at  a  distance  of  2  ft.  from  its 
mean  position.    Find  the  amplitude  of  the  motion. 

31.  A  particle  moves  according  to  the  equation  s  =  4  sin  Jy  ^^  + 
5  cos  \  t.  Show  that  the  motion  is  simple  harmonic  and  find  the 
amplitude  of  the  swing  and  the  time  at  which  the  particle  passes 
through  its  mean  position. 

32.  Find  the  radius  of  curvature  of  the  curve  y  =  x  sin  -  at  the 

2  ^ 

point  for  which  x  —  — 

TT 

sin  X 

33.  Find  the  radius  of  curvature  of  the  curve  //  = at  the 

X 

point  for  which  ,t  =  tt. 

X  I 

34.  Find  the  radius  of  curvature  of  the  curve  y  =  a  sin~^  -  —  ^d^  —  x^ 

a  ^ 

at  the  point  for  which  x  =^  -• 


35.  Find  the  radius  of  curvature  of  the  curve  x  =  a  cos  <f},  y  =  b  sin  cf), 

TT 

at  the  point  for  which  cf>  =  —- 

Plot  the  graphs  of  the  following  curves  : 

36.  1^  =  a"^ sin -•  41.   r  =1  —  26. 

42.  7-2  =4  sec  2^. 

37.  r2  =  a^ sin  4:0.  .^      2        2^      a 

43.  r^=  a^ta,nO. 

38.  r  =  a(l  —  sin  6).  $ 

^  ^  44.   /•  =l  +  sin-. 

39.  r  =a(l  +cos2^).  ^ 

40.  r  =  ^t(l  +  2  sin  6).  45.   r  =  1  -f-  sin  -7-- 

Find  the  points  of  intersection  of  the  following  pairs  of  curves : 

46.  A-2  =  3  cos  2  6,  t^  =  2  cos"  6. 

47 .  r  =  a  cos  6,  r^  =  a^  sin  2  B. 

48.  /•  =  2  sin  e,  ?-2  =  4  sin  2  B. 

49.  r  =  «(1  +  sin  2  B),  r^  =  4  a^  sin  2  B. 


GENERAL  EXERCISES  151 

Transform  the  following  curves  to  polar  coordinates : 

50.    1/  = .  51.   v/  +  7/V-^  -  ^V  ==  0. 

Transform  the  following  curves  to  .xy-coordinates  : 

52.   /•-  =  2  a'  sin  2  0.  53.   /•  =  a  (1  —  cos  6). 

54.  Find  the  angle  at  which  the  curve  r  =  3  -f-  sin  2  6  meets  the 
circle  r  =  3. 

55.  Find  the  angle  of  intersection  of  the  two  curves  r  =  2  sin  d 

and  t^  =  4:  sin  2  0. 

56.  Find  the  angle  of  intersection  of  the  curves  r  =  a  cos  0  and 
7'  =  a  sin  2  $. 

57.  If  a  ball  is  fired  from  a  gun  with  the  initial  velocity  v^,  it 

QX 

describes  a  path  the  equation  of  which  is  y  =  xtsma  —  - — r --) 

^  ^  "^  2?;(fcosvj: 

where  a  is  the  angle  of  elevation  of  the  gun  and  OX  is  horizontal. 

What  is  the  value  of  a  when  the  horizontal  range  is  greatest  ? 

58.  In  measuring  an  electric  current  by  means  of  a  tangent  galva- 
nometer, the  percentage  of  error  due  to  a  small  error  in  reading  is 
proportional  to  tan  x  -+-  ctn  x.  For  what  value  of  x  will  this  percent- 
age of  error  be  least  ? 

59.  A  tablet  8  ft.  high  is  placed  on  a  wall  so  that  the  bottbm  of 
the  tablet  is  29  ft.  from  the  ground.  How  far  from  the  wall  should 
a  person  stand  in  order  that  he  may  see  the  tablet  to  best  advantage 
(that  is,  that  the  angle  between  the  lines  from  his  eye  to  the  top  and 
to  the  bottom  of  the  tablet  should  be  the  greatest),  assuming  that 
liis  eye  is  5  ft.  from  the  ground  ? 

60.  One  side  of  a  triangle  is  12  ft.  and  the  opposite  angle  is  36°. 
Find  the  other  angles  of  the  triangle  when  its  area  is  a  maximum. 

61.  Above  the  center  of  a  round  table  of  radius  2  ft.  is  a  hanging 
lamp.  How  far  should  the  lamp  be  above  the  table  in  order  that 
the  edge  of  the  table  may  be  most  brilliantly  lighted,  given  that 
the  illumination  varies  inversely  as  the  square  of  the  distance  and 
directly  as  the  cosine  of  the  angle  of  incidence  ? 

62.  A  weight  P  is  dragged  along  the  ground  by  a  force  F.  If 
the  coefficient  of  friction  is  k,  in  what  direction  should  the  force  be 
applied  to  produce  the  best  result  ? 


152  TRIGONOMETRIC  FUNCTIONS 

63.  An  open  gutter  is  to  be  constructed  of  boards  in  such  a  way 
that  the  bottom  and  sides,  measured  on  the  inside,  are  to  be  each 
6  in.  wide  and  both  sides  are  to  have  the  same  slope.  How  wide 
should  the  gutter  be  across  the  top  in  order  that  its  capacity  may 
be  as  great  as  possible  ? 

64.  A  steel  girder  27  ft.  long  is  to  be  moved  on  rollers  along  a 
passageway  and  into  a  corridor  8  ft.  in  width  at  right  angles  to  the 
passageway.  If  the  horizontal  width  of  the  girder  is  neglected,  how 
wide  must  the  passageway  be  in  order  that  the  girder  may  go  around 
the  corner  ? 

65.  Two  particles  are  moving  in  the  same  straight  line  so  that 
their  distances  from  a  fixed  point  0  are  respectively  x  =  a  cos  kt  and 

a;'  =  acoslH  4- — ),  k  and  a  being  constants.    Find  the  greatest 

distance  between  them. 

66.  Show  that  for  any  curve  in  polar  coordinates  the  maximum 
and  the  minimum  values  of  r  occur  in  general  when  the  radius  vector 
is  perpendicular  to  the  curve. 

67.  Two  men  are  at  one  end  of  the  diameter  of  a  circle  of  40  yd. 
radius.  One  goes  directly  toward  the  center  of  the  circle  at  the 
uniform  rate  of  6  ft.  per  second,  and  the  other  goes  around  the 
circumference  at  the  rate  of  2  tt  ft.  per  second.  How  fast  are  they 
separating  at  the  end  of  10  sec.  ? 

68.  Given  that  two  sides  and  the  included  angle  of  a  triangle  are 
6  ft.,  10  ft.,  and  30°  respectively,  and  are  changing  at  the  rates  of 
4  ft.,  —  3  ft.,  and  12°  per  second  respectively,  what  is  the  area  of  the 
triangle  and  how  fast  is  it  changing  ? 

69.  A  revolving  light  in  a  lighthouse  \  mi.  offshore  makes  one 
revolution  a  minute.  If  the  line  of  the  shore  is  a  straight  line,  how 
fast  is  the  ray  of  light  moving  along  the  shore  when  it  passes  a 
point  one  mile  from  the  point  nearest  to  the  lighthouse  ? 

70.  -BC  is  a  rod  a  feet  long,  connected  with  a  piston  rod  at  C,  and 
at  B  with  a  crank  AB,  h  feet  long,  revolving  about  A.  Find  C's 
velocity  in  terms  of  .45's  angular  velocity. 

71.  At  any  time  t  the  coordinates  of  a  point  moving  in  the  ic^z-plane 
are  x  =  2  —  3  cos  t,  y  =  3  -\-  2  sin  t.  Find  its  path  and  its  velocity  in 
its  path.    At  what  points  will  it  have  a  maximum  velocity  ? 

72 .  At  any  time  t  the  coordinates  of  a  moving  point  are  x  =  2  sec  3  /, 
?y  =  4  tan  3  f.   Find  the  equation  of  its  path  and  its  velocity  in  its  path. 


I 


GENERAL  EXERCISES  153 


73.  The  parametric  equations  of  the  path  of  a  moving  particle  are 
x  =  2cos^<f>,  y  =  2sin^<l>.  If  the  angle  <f>  increases  at  the  rate  of 
2  radians  per  second,  find  the  velocity  of  the  particle  in  its  path. 

74.  A  particle  moves  along  the  curve  ?/  =  sina;  so  that  the 
£c-component  of  its  velocity  has  always  the  constant  value  a.  Find 
the  velocity  of  the  particle  along  the  curve  and  determine  the  points 
of  the  curve  at  which  the  particle  is  moving  fastest  and  those  at 
which  it  is  moving  most  slowly. 

75.  Find  the  angle  of  intersection  of  the  curves  2/  =  sin  a;  and 
y  =  cos  X. 

76.  Find  the  angle  of  intersection  of  the  curves  y  =  sina;  and 
y  =  smlx  +  ^y 

77.  Find  the  angle  of  intersection  of  the  curves  ?/  =  sina;  and 
y  =  cos  2  X  between  the  lines  x  =  0  and  a;  =  2  tt. 

78.  Find  the  points  of  intersection  of  the  curves  y  =  sin  a;  and 
y  =  sin  3  x  between  the  lines  x  =  0  and  x  =  7r.  Determine  the  angles 
at  the  points  of  intersection. 

79.  Find  all  the  points  of  intersection  of  the  curves  y  =  cosx  and 
2/  =  sin  2  ic  which  lie  between  the  lines  x  =  0  and  cc  =  2  tt,  and 
determine  the  angles  of  intersection  at  each  of  the  points  found. 


CHAPTER  VI 
EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

53.  The  exponential  function.    The  equation 

y  =  «^ 

where  a  is  any  constant,  defines  ^  as  a  function  of  x  called  the 

exponential  function. 

If  x  =  n,  an  integer,  j/  is  determined  by  raising  a  to  the  nth 

power  by  multiplication. 

p 
If  x  =  -i  a  positive  fraction,  y  is  the  ^-th  root  of  the  pih 

power  of  a. 

If  a:  is  a  positive  irrational  number,  the  approximate  value  of 
2/  may  be  obtained  by  expressing  x  approximately  as  a  fraction. 

li  x=  0,  1/  =  a^  =  l.    If  a;  =  —  w,  y  =  a ~ "*  =  — . 
The  graph  of  the  function  is  readily  found. 

Ex.  Find  the  graph  oi  y  =  (1.5)\  By  giving  convenient  values  to  .r 
we  obtain  the  curve  shown  in  Fig.  70.  To  determine  the  shape  of  the 
curve  at  the  extreme  left,  we  place  x  equal  to  a  large  negative  number, 

say  x=-  100.    Then  y  =  (1.5)-ioo  = 1 — , 

J  if      K      J  (1.5)100 

which  is  very  small.  It  is  obvious  that  the 
larger  numerically  the  negative  value  of  x 
becomes,  the  smaller  y  becomes,  so  that  the 
curve  approaches  asymptotically  the  negative 
portion  of  the  a;-axis. 

On  the  other  hand,  if  a:  is  a  large  positive 
number,  y  is  also  large.  Fig.  70 

54.  The  logarithm.  If  a  number  JS/"  may  be  obtained  by  placing 
an  exponent  L  on  another  number  a  and  computing  the  result, 
then  L  is  said  to  be  the  logarithm  of  N  to  the  base  a.    That  is,  if 

.    2^=  «^  (1) 

then  L  =  \og^]Sr.  (2) 

154 


LOGARITHMS  155 

Formulas  (1)  and  (2)  are  simply  two  different  ways  of  ex- 
pressing the  same  fact  as  to  the  relation  of  N  and  X,  and  the 
student  should  accustom  himself  to  pass  from  one  to  the  other 
as  convenience  may  demand. 

From  these  formulas  follow  easily  the  fundamental  properties 
of  logarithms;  namely, 

log«^+  log„  Jf  =  log„  JfiV, 

log„i\^-log„Jf=log,-, 

n\og,N=log„N\  (3) 

logj=0, 

Theoretically  any  number,  except  0  or  1,  may  be  used  as 
the  base  of  a  system  of  logarithms.  Practically  there  are  only 
two  numbers  so  used.  The  first  is  the  number  10,  the  use  of 
which  as  a  base  gives  the  common  system  of  logarithms,  which 
are  the  most  convenient  for  calculations  and  are  used  almost 
exclusively  in  trigonometry. 

Another  number,  however,  is  more  convenient  in  theoretical 
discussions,  since  it  gives  simpler  formulas.  This  number  is 
denoted  by  the  letter  e  and  is  expressed  by  the  infinite  series 

,1111 

'==^"^1^21  +  3!  ■^^^•••' 

where  2  !  =1  x  2,  3  !  =1  x  2  x  3,  4  !  =1  x  2  x  3  x  4,  etc. 

Computing  the  above  series  to  seven  decimal  places,  we  have 

^  =  2.7182818.... 

An  important  property  of  this  number,  which  is  necessary  in 
finding  the  derivative  of  a  logarithm,  is  that 


156    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

To  check  this  arithmetically  we  may  take  successive  small 
values  of  h  and  make  the  following  computation : 

When  A=.l,  (1  + A)*=  (1.1)^°=  2.59374. 

When  ^=.01,        (1-f- A)^=(1.01)^°°=  2.70481. 

When  h  =  .001,      (1  +  hy=  (1.001)^°°^=  2.71692. 

When  A  =.0001,    (1+ A)^=  (1.0001)^'^°'^'^=  2.71815. 

1 
Working  algebraically,  we  expand  (1-j-hy  by  the  binomial 

theorem,  obtaining        ^  lA-lVl-2\ 

1      (1-h)      (1-A)(1-2A) 
12!  3! 

where  E  represents  the  sum  of  all  terms  involving  A,  h%  h^,  etc. 
Now  it  may  be  shown  by  advanced  methods  that  as  h  approaches 
zero,  H  also  approaches  zero ;  so  that 

L™(l+A)^=l+l+i,  +  l  +  ...  =  . 

When  the  number  e  is  used  as  the  base  of  a  system  of  loga- 
rithms, the  logarithms  are  called  natural  logarithms,  or  Napierian 
logarithms.  We  shall  denote  a  natural  logarithm  by  the  symbol 
In*  ;  thus, 

if  N=e\       . 

(4) 
then  X  =  lniV; 

Tables  of  natural  logarithms  exist,  and  should  be  used  if 
possible.    In  case  such  a  table  is  not  available,  the  student 

*  This  notation  is  generally  used  by  engineers.  The  student  should  know 
that  the  abbreviation  "log"  is  used  by  many  authors  to  denote  the  natural 
logarithm.    In  this  book  "  log  "  is  used  for  the  logarithm  to  the  base  10. 


LOGARITHMS 


157 


may  find  the  natural  logarithm  by  use  of  a  table  of  common 
logarithms,  as  follows: 

Let  it  be  required  to  find  In  213. 

If  :r  =  ln213, 

then,  by  (4),  213  =  e^; 

whence,  by  (3),  log  21S  =  x  log  e, 

log  213        2.3284 


or 


log2.7183      0.4343 


5.36L 


Certain  graphs  involving  the  number    y 
e  are  important  and  are  shown  in  the 
examples. 

Ex.  1.   !/  =  Inx. 

Giving  X  positive  values  and  finding  ij,  we 
obtain  Fig.  71. 


Fig.  71 


y 

The  curve  (Fig.  72)  is  symmetrical  vrith  respect  to  0  F  and  is  always 
above  OX.  When  x  =  0,  y  =  1.  As  a:  increases  numerically,  y  decreases, 
approaching  zero.    Hence  OX  is  an  asymptote. 


Fig.  72 


Fig.  73 


Ex. 


t.    y  =  _(e«4.e    «). 


This  is  the  curve  (Fig.  73)  made  by  a  cord  or  a  chain  held  at  the  ends 
and  allowed  to  hang  freely.    It  is  called  the  catenary. 

Ex.  4.    2/  =  e-"^sinbx. 

The  values  of  y  may  be  computed  by  multiplying  the  ordinates  of  the 
curve  y  =  e-'^^hj  the  values  of  sin  hx  for  the  corresponding  abscissas.  Since 
the  value  of  smhx  oscillates  between  1  and  —1,  the  values  of  g-"^ sin 6a: 


158    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 


cannot  exceed  those  of  e~* 
plane  between  the  curves  y  =  e-^^ 
and  y=  —  e~^.    When  x  is  a  mul- 
tiple of  -,  y  is  zero.    The  graph 

therefore  crosses  the  axis  of  x  an 
infinite  number  of  times.    Fig.  74 
shows  the  graph  when  a  =  1,  &  =  2  tt. 
1 

Ex.  b.    y  =  e^. 

When  X  approaches  zero,  being 
positive,  y  increases  without  limit. 
When  X  approaches  zero,  being  neg- 
ative, y  approaches  zero  ;  for  example,  when 
^  =  TTJ^iya»  y  =  ^^°°^  ^^^  when  x  =  -  roVu* 
y  =  e  -"^^^^  = •   The  function  is  therefore 

^  glOOO 

discontinuous  for  x  =  0. 

The  line  r/  =  1  is  an  asymptote  (Fig.  75), 
for  as  X  increases  without  limit,  being  posi- 
tive or  negative,  —  approaches  0,  and  y 
approaches  1. 

Ex.  6.    r  =  e«^. 

The  use  of  r  and  6  indicates  that  we  are 
using  polar  coordinates. 

When  ^  =  0,  r  =  1.  As  0  increases,  r  in- 
creases, and  the  curve  winds  around  the  origin 
at  increasing  distances  from  it  (Fig.  76). 
When  6  is  negative  and  increasing  numer- 
ically without  limit,  r  approaches  zero. 
Hence  the  curve  winds  an  infinite  number 
of  times  around  the  origin,  continually  ap- 
proaching it.  The  dotted  line  in  the  figure 
corresponds  to  negative  values  of  6. 

The  curve  is  called  the  logarithmic  spiral. 


Hence  the  graph  lies  in  the  portion  of  the 
Y 


Tig.  75 


Fig.  76 


EXERCISES 

Plot  the  graphs  of  the  following  equations  : 
1.  y  =  (1)-.  h,  y  =  xe\ 

6.  y  =  \{e^-e-% 

7.  7/  =  log  1x. 

,    1 

8.  ]/  =  log-- 


2.  y  =  (^y. 

3.  y  =  e^  +  i 

4 


y 


e 


9.  y  =  log  since. 

10.  y  =  log  tan  ic. 

11.  y  =  e~^^sin  4,x. 

12.  y  =  e"^cos  Sx. 

13.  r=e-2^ 


EMPIRICAL  EQUATIONS 


159 


55.  Certain  empirical  equations.  If  x  and  y  are  two  related 
quantities  which  are  connected  by  a  given  equation,  we  may 
plot  the  corresponding  curve  on  a  system  of  a;^-coordinates,  and 
every  point  of  this  curve  determines  corresponding  values  of 
X  and  y. 

Conversely,  let  x  and  y  be  two  related  quantities  of  which 
some  corresponding  pairs  of  values  have  been  determined,  and 
let  it  be  desired  to  find  by  means  of  these  data  an  equation  con- 
necting X  and  y  in  general.  On  this  basis  alone  the  problem 
cannot  be  solved  exactly.  The  best  we  can  do  is  to  assume  that 
the  desired  equation  is  of  a  certain  form  and  then  endeavor  to 
adjust  the  constants  in  the  equation  in  such  a  way  that  it  fits 
the  data  as  nearly  as  possible.    We  may  proceed  as  follows : 

Plot  the  points  corresponding  to  the  known  values  of  x  and  y. 
The  simplest  case  is  that  in  which  the  plotted  points  appear  to 
lie  on  a  straight  line  or  nearly  so.  In  that  case  it  is  assumed 
that  the  required  relation  may  be  put  in  the  form 

y  =  mx-\-b,  (1) 

where  m  and  h  are  constants  to  be  determined  to  fit  the  data. 
The  next  step  is  to  draw  a  straight  line  so  that  the  plotted 
points  either  lie  on  it  or  are  close  to  it  and  about  evenly  dis- 
tributed on  both  sides  of  it.  The  equation  of  this  line  may  be 
found  by  means  of  two  points  on  it,  which  may  be  either  two 
points  determined  by  the  original  data  or  any  other  two  points 
on  the  Ime. 

The  resulting  equation  is  called  an  empirical  equation  and 
expresses  approximately  the  general  relation  between  x  and  y. 
In  fact,  more  than  one  such  equation  may  be  derived  from  the 
same  data,  and  the  choice  of  the  best  equation  depends  on  the 
judgment  and  experience  of  the  worker. 

Ex.  1.  Corresponding  values  of  two  related  quantities  x  and  y  are  given 
by  the  following  table  : 


X 

1 

2 

4 

6 

10 

y 

1.3 

2.2 

2.9 

3.9 

6.1 

Find  the  empirical  equation  connecting  them. 


160    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 


We  plot  the  points  (x,  rj)  and  draw  the  straight  line,  as  shown  in 
Fig.  77.  The  straight  line  is  seen  to  pass  through  the  points  (0,  1)  and 
(2,  2).    Its  equation  is  therefore 


y  =  .5  ar  +  1, 
which  is  the  required  equation. 


In  many  cases,  however, 
the  plotted  points  will  not 
appear  to  lie  on  or  near  a 


straight  line.  We  shall  con-  ^ 
sider  here  only  two  of  these 
cases,  which  are  closely  con- 
nected with  the  case  just 
considered.  They  are  the  cases  in  which  it  may  be  anticipated 
from  previous  experience  that  the  required  relation  is  either 
of  the  form 


Fig.  77 


y  =  ah'', 
where  a  and  h  are  constants,  or  of  the  form 


(2) 


(3) 


y  =  ax-\ 

where  a  and  n  are  constants. 

Both  of  these  cases  may  be  brought  directly  under  the  first 
case  by  taking  the  logarithm  of  the  equation  as  written.  Equa- 
tion (2)  then  becomes 

log  y  =  log  a-{-x  log  h.  (4) 

As  log  a  and  log  h  are  constants,  if  we  denote  log  y  by  y\ 
(4)  assumes  the  form  (1)  in  x  and  ?/',  and  we  have  only  to 
plot  the  points  (x,  ?/')  on  an  xy'-^y^tQm.  of  axes  and  determine 
a  straight  line  by  means  of  them.  The  transformation  from  (4) 
back  to  (2)  is  easy,  as  shown  in  Ex.  2. 

Taking  the  logarithm  of  (3),  we  have 

log  y  =  log  a-\-n  log  x.  (5) 

If  we  denote  log  y  by  y'  and  log  x  by  x\  (5)  assumes  the  form 
(1)  in  x'  and  y',  since  log  «  and  ti  are  constants.  Accordingly 
we  plot  the  points  (x',  y)  on  an  x' y' -sy^tQVLi  of  axes,  determine 
the  corresponding  straight  line,  and  then  transform  back  to  (3), 
as  shown  in  Ex.  3. 


EMPIRICAL  EQUATIONS 


161 


Ex,  2.    Corresponding  values  of  two  related  quantities  x  and  y  are  given 
by  the  following  table  : 


X 

8 

10     12 

14 

16 

18 

20 

V 

3.2 

4.6 

7.3 

9.8 

15.2 

24.6 

36.4 

,_  ., 

Find  an  empirical  equation  of  the  form  y  =  alF. 

Taking  the  logarithm  of  the  equation  y  =  aZ/*",  and  denoting  log  y  by  y\ 

we  have 

y'  =  log  a  ■\-  X  log  h. 

Determining  the  logarithm  of  each  of  the  given  values  of  y,  we  form  a 
table  of  corresponding  values  of  x  and  y',  as  follows : 


X 

8 

10 

12 

14 

16 

18 

20 

y'  =  \ogy 

.5051 

.6628 

.8633 

.9912 

1.1818 

1.3909 

1.5611 

We  choose  a  large-scale  plotting-paper,  assume  on  the  ?/'-axis  a  scale 
four  times  as  large  as  that  on  the 
X-axis,  plot  the  points  (x,  /),  and 
draw   the   straight   line    (Fig.  78)     2.0 
through   the    first    and    the    sixth     1.5 

1.0 
.5 


point.    Its  equation  is 

v'  =  . 08858 a: -.20354. 


2    4    6    8    10  12  14  16  18  20 
Fig.  78 


Therefore     log  a  =  -  .20354  = 

9.7965  - 10,  whence  a  =  .626  ;  and 

log  b  =  .08858,  whence  h  =  1.22.    Substituting  these  values  in  the  assumed 

equation,  we  have  ^^^  ^  ^^ 

^  '  y  =  .626  (1.22)^ 

as  the  required  empirical  equation.  The  result  may  be  tested  by  substitut- 
ing the  given  values  of  x  in  the  equation.  The  computed  values  of  y  will 
be  found  to  agree  fairly  well  with  the  given  values. 

Ex.  3.  Corresponding  values  of  pressure  and  volume  taken  from  an 
indicator  card  of  an  air -compressor  are  as  follows : 


p 

18 

21 

26.5 

33.5 

44 

62 

V 

.635 

.556 

.475  1  .397 

.321 

.243 

Find  the  relation  between  them  in  the  form  pv** 


162    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

Writing  the  assumed  relation  in  the  form  p  =  cv-^  and  taking  the 
logarithms  of  both  sides  of  the  equation,  we  have 

\ogp  =—  n  log  V  +  log  c, 
or  y  =—  nx  +  b, 

"where  y  =  log/>,  x  =  log  v,  and  h  =  log  c. 

The  corresponding  values  of  x  and  y  are 


a;  =  log  u 

-  J972 

-  .2549 

-  .3233 

-  .4012 

-  .4935 

-  .6144 

y  =  logp 

1.2553 

1.3222 

1.4232 

1.5250         1.6435 

1.7924 

For  convenience  we  assume  on  the  a:-axis  a  scale  twice  as  large  as  that 

the  3/-axis,  plot  the  points  (x,  ?/),  and  draw  the  straight  line  as  shown 

Fig.  79.  The  construction  should 

be  made  on  large-scale  plotting- 

paper.    The  line  is  seen  to  pass 

through  the  points  (—  .05, 1.075) 

and  (—  .46, 1.6).   Its  equation  is 

therefore 

1/ =- 1.28  x  + 1.01. 

Hence  n  =  1.28,  logc  =  1.01, 
c  =:10.2,  and  the  required  rela- 
tion between  p  and  v  is 
,1-28  =  10.2. 


on 
in 


pv^ 


-5o-50-k5-l*O-35-.30-25-20':15-10'.O5  0 

Fig.  79 


EXERCISES 


1.  Show  that  the  following  points  lie  approximately  on  a  straight 
line,  and  find  its  equation  : 


X 

4 

9 

13 

20 

22 

25 

30 

y 

2.1 

4.6 

7 

12 

12.9 

14.5 

18.2 

2.  For  a  galvanometer  the  deflection  D,  measured  in  millimeters 
on  a  proper  scale,  and  the  current  7,  measured  in  microamperes,  are 
determined  in  a  series  of  readings  as  follows  : 


D 

29.1 

48.2 

72.7 

92.0 

118.0 

140.0 

165.0 

199.0 

I 

0.0493 

0.0821       0.123 

0.154 

0.197 

0.234 

0.274 

0.328 

Find  an  empirical  law  connecting  D  and  /. 


I 


EMPIRICAL  EQUATIONS 


163 


3.  Corresponding  values  of  two  related  quantities  x  and  y  are 
given  in  the  following  table  : 


X 

0.1 

0.3 

0.5 

0.7 

0.9 

1.1 

1.3 

1.6 

V 

0.3316 

0.4050 

0.4946 

0.6041 

0.7379 

0.9013 

1.1008 

1.3446 

Find  an  empirical  equation  connecting  x  and  y  in  the  form  y  =  ah^. 

4.  In  a  certain  chemical  reaction  the  concentration  c  of  sodium 
acetate  produced  at  the  end  of  the  stated  number  of  minutes  t  is 
as  follows  : 


t 

1 

2 

3 

4 

5 

c 

0.00837 

0.00700 

0.00586 

0.00492 

0.00410 

Eind  an  empirical  equation  connecting  c  and  t  in  the  form  c  =  ab\ 
5.  The  deflection  a  of  a  loaded  beam  with  a  constant  load  is 
found  for  various  lengths  I  as  follows  : 


I 

1000 

900 

800 

700 

600 

a 

7.14 

5.22 

3.64 

2.42 

1.50 

Eind  an  empirical  equation  connecting  a  and  I  in  the  form  a  =  nl^. 
6.  The  relation  between  the  pressure  p  and  the  volume  v  of  a  gas 
is  found  experimentally  as  follows : 


p 

20 

23.5 

31 

42 

59 

78 

V 

0.619 

0.540 

0.442 

0.358 

0.277 

0.219 

Eind  an  empirical  equation  connecting  p  and  v  in  the  form/^y"  =  c. 

56.  Differentiation.  The  formulas  for  the  differentiation  of 
the  exponential  and  the  logarithmic  functions  are  as  follovi^s, 
where,  as  usual,  u  represents  any  function  which  can  be  differen- 
tiated with  respect  to  x,  In  means  the  Napierian  logarithm,  and 
a  is  any  constant: 

rf  Inor    />  rft/ 


d  ,  loff„  e  du 

dx     ^"  u      dx 


d  ^       _\  du 
dx  u  dx 


(2) 


164    EXPONENTIAL  AND  LOGAEITHMIC  FUNCTIONS 


(4) 


dx  dx 

—  6«  =  e«  — ^ 
dx  dx 

The  proofs  of  these  formulas  are  as  follows: 
l.By(8>§36,£log„.  =  |-log.».|. 
To  find  —  log„w  place  y=  log^u. 

Then,  if  u  is  given  an  increment  Am,  y  receives  an  increment 
Ai/,  where 

^y  =  log«(^  +  Aw)  -  log^w 
the  transformations  being  made  by  (3),  §  54. 


Au 


Then  ^=liogil  +  42f 

Aw      u     ^  \         u 


XAm 


Aw 
Now,  as  Aw  approaches  zero  the  fraction  —  may  be  taken 

as  ^  of  §  54.  ^ 

/       Aw\^ 
Hence  Lim  ( 1 H )    =  e. 

Therefore  ^  =  lw  g 

du     u    ^" 

and  ^^log^^^ 

dx         u     dx 

2.  If  ?/  =  In  w,  the  base  a  of  the  previous  formula  is  e ;  and 
since  log^  e=l,  we  have 

dy  _1  du 
dx      u  dx 


DIFFERENTIATION  165 


3.  If  ^  =  «", 

we  have  In^  =  In  a"  =  m  In  a. 

Hence,  by  formula  (2), 

1  dy  _^       du  ^ 

y  dx  dx 

dy       „ ,       du 
whence  -^  =  a"  In  a  -— • 

dx  dx 

4.  li  y  =  ^"  the  previous  formula  becomes 


dy_ 
dx 

'1- 

Ex. 

1. 

y  = 

ln(a:2- 

4a: 

+ 

5). 
dy_ 

2x- 

-  4 

dx 

x2-4x  +  5 

Ex. 

2. 

y  = 

e-\ 

dy  _ 
dx 

-  2a:e- 

■< 

Ex. 

3. 

y  = 

:  e-«^cos6ar. 

-^  =  COS  hx  —  {e-  «^)  +  e-  "^  —  (cos  6a:)  =  —  ae-  "^  cos  6a:  —  he-  «^sin  6a: 
^^  ^^  ^-^  =-c-«^(acos6a:  +  6sin6x). 

EXERCISES 

Find  -r  in  each  of  the  following  cases : 
dx 

.    1  —  sin  2x 

10.  2/  =  In:; r— ^^ — 

^  1  +  sm  2  ic 

11.  ?/  =  ln(e2^+e-2^). 

12.  y  —  e-2*sin  3  a-. 


1. 

2/  = 

e  ^. 

2. 

2/  = 

:K^"+^"')- 

3. 

2/  = 

:  a^-\ 

4. 

?/  = 

:  a"""X 

5. 

2/  = 

:ln(a:2+4a!- 

■1> 

6. 

:lnV2x2+6 

ic  +  9. 

7. 

2/  = 

=  ^^%  +  3- 

13.  2/  =  In  Vl4-ic^  +  a:ctn-^a!. 

14.  2/  =  e«^(9a;2_  6a; -f  2). 

15.  2/  =  e2^(2  sin  a;  —  cos  a;). 


8.  2/  =  In (x  +  Va;^+4). 

9.  2/  =  In  (3a3  +  VOa^^  +  l).  16.  2/  =  sin     ^^  _^  ^ 

17.  2/ =  seca:tancc  4- ln(seca;  +  tanx). 

,     V^Tl-1 

18.  y  =  In  -    .  -• 


166    EXPONENTIAL  AND  LOGAKITHMIC  FUNCTIONS 

57.  The  compound-interest  law.  An  important  use  of  the  ex- 
ponential function  occurs  in  the  problem  to  determine  a  function 
whose  rate  of  change  is  proportional  to  the  value  of  the  function. 
If  y  is  such  a  function  of  x^  it  must  satisfy  the  equation 

|=%.  m 

where  A:  is  a  constant  called  the  proportionality  factor. 
We  may  write  equation  (1)  in  the  form 

y  dx 

whence,  by  a  very  obvious  reversal  of  formula  (2),  §  56,  we  have 

In  y  =  kx-\-C, 

where  C  is  the  constant  of  integration  (§18). 
From  this,  by  (1)  and  (2),  §  54, 


Finally  we  place  e^=A,  where  A  may  be  any  constant,  since 
C  is  any  constant,  and  have  as  a  final  result 

•     y=AeK  (2) 

The  constants  A  and  h  must  be  determined  by  other  condi- 
tions of  a  particular  problem,  as  was  done  in  §  18. 

The  law  of  change  here  discussed  is  often  called  the  compound- 
interest  law,  because  of  its  occurrence  in  the  following  problem : 

Ex.   Let  a  sum  of  money  P  be  put  at  interest  at  the  rate  of  r%  per  annum. 

The  interest  gained  in  a  time  A^  is  P A^  where  A^  is  expressed  in 

years.  But  the  interest  is  an  increment  of  the  principal  P,  so  that  we  have 

AP=P-^A/. 
100 

In  ordinary  compound  interest  the  interest  is  computed  for  a  certain 
interval  (usually  one-half  year),  the  principal  remaining  constant  during 
that  interval.  The  interest  at  the  end  of  the  half  year  is  then  added  to  the 
principal  to  make  a  new  principal  on  which  interest  is  computed  for  the 


COMPOUND-INTEREST  LAW  167 

next  half  year.   The  principal  P  therefore  changes  abruptly  at  the  end  of 
each  half  year. 

Let  us  now  suppose  that  the  principal  changes  continuously ;  that  is, 
that  any  amount  of  interest  theoretically  earned,  in  no  matter  how  small 
a  time,  is  immediately  added  to  the  principal.  The  average  rate  of  change 
of  the  principal  in  the  period  A<  is,  from  §  11, 

AP       Pr 

M       100  ^  ^ 

To  obtain  the  rate  of  change  we  must  let  A/  approach  zero  in  equation 
(1),  and  have  ,_, 

^—  p  -L-. 
dr~      lOO' 

From  this,  as  in  the  text,  we  have 

P=Ae^o\  (2) 

To  make  the  problem  concrete,  suppose  the  original  principal  were  $100 
and  the  rate  4%,  and  we  ask  what  would  be  the  principal  at  the  end  of  14  yr. 
We  know  that  when  <  =  0,  P  =  100.  Substituting  these  values  in  (2),  we 
have  A  =  100,  so  that  (2)  becomes 

P  =  100eioo  =100e25. 

Placing  now  i  =  14,  we  have  to  compute  P  =  100  e^'S.  The  value  of  P 
may  be  taken  from  a  table  if  the  student  has  access  to  tables  of  powers  of  e. 
In  case  a  table  of  common  logarithms  is  alone  available,  P  may  be  found 
by  first  taking  the  logarithm  of  both  sides  of  the  last  equation.    Thus 

log  P  =  log  100  +  it  log  e  =  2.4053  ; 
whence  P  =  $254,  approximately. 

EXERCISES 

1.  The  rate  of  change  of  y  with  respect  to  x  is  always  equal  to 
^  y,  and  when  x  =  0,  y  =  5.    Find  the  law  connecting  y  and  x. 

2.  The  rate  of  change  of  y  with  respect  to  x  is  always  0.01  times  y, 
and  when  x  =  10,  y  =  50.    Find  the  law  connecting  y  and  x. 

3.  The  rate  of  change  of  y  with  respect  to  x  is  proportional  to  y. 
When  x  =  0,  y  =  7,  and  when  x  =  2,  ?/  =  14.  Find  the  law  connect- 
ing y  and  X. 

4.  The  sum  of  1 100  is  put  at  interest  at  the  rate  of  5%  per  annum 
under  the  condition  that  the  interest  shall  be  compounded  at  each 
instant  of  time.    How  much  will  it  amount  to  in  40  yr.  ? 


168    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

5.  At  a  certain  date  the  population  of  a  town  is  10,000.  Forty 
years  later  it  is  25,000.  If  the  population  increases  at  a  rate  which 
is  always  proportional  to  the  population  at  the  time,  find  a  general 
expression  for  the  population  at  any  time  t. 

6.  In  a  chemical  reaction  the  rate  of  change  of  concentration  of 
a  substance  is  proportional  to  the  concentration  at  any  time.  If  the 
concentration  is  y^^  when  ^  =  0,  and  is  ^^3^  when  t  =  h^  find  the  law 
connecting  the  concentration  and  the  time. 

7.  A  rotating  wheel  is  slowing  down  in  such  a  manner  that  the 
angular  acceleration  is  proportional  to  the  angular  velocity.  If  the 
angular  velocity  at  the  beginning  of  the  slowing  down  is  100  revolu- 
tions per  second,  and  in  1  min.  it  is  cut  down  to  50  revolutions  per 
second,  how  long  will  it  take  to  reduce  the  velocity  to  25  revolutions 
per  second  ? 

GENERAL  EXERCISES 

Plot  the  graphs  of  the  following  equations : 


1-  y  =  (i)- 

2.  2/=  e>-» 

X 
~2 


3.  y 


e    "COS  a:!. 


4.  ?/  =  e^""*". 

5.  2/=i(^"+e"")- 


6-  2/  =  :t 


e^+^' 


7.  y  =  xe^. 


9.  y  =  x 


—  ^2„^a; 


10.  For  a  copper-nickel  thermocouple  the  relation  between  the 
temperature  t  in  degrees  and  the  thermoelectric  power  p  in  micro- 
volts is  given  by  the  following  table : 


t 

0 

50 

100 

150 

200 

p 

24 

25 

26 

26.9 

27.5 

Find  an  empirical  law  connecting  t  and  p. 

11.  The  safe  loads  in  thousands  of  pounds  for  beams  of  the  same 
cross  section  but  of  various  lengths  in  feet  are  found  as  follows  : 


Length 

10 

11 

12 

13 

14 

15 

Load 

123.6 

121.5 

111.8 

107.2 

101.3 

90.4 

Find  an  empirical  equation  connecting  the  data. 


GENERAL  EXERCISES 


169 


12.  In  the  following  table  s  denotes  the  distance  of  a  moving 
body  from  a  fixed  point  in  its  path  at  time  t : 


t 

1 

2 

4 

6 

7 

8 

s 

10 

4 

0.6400 

0.1024 

0.0410 

0.0164 

Find  an  empirical  equation  connecting  s  and  t  in  the  form  s  =  abK 

13.  In  the  following  table  c  denotes  the  chemical  concentration  of 
a  substance  at  the  time  t : 


t 

2 

4 

6 

8 

10 

c 

0.0069 

0.0048 

0.0033 

0.0023 

0.0016 

Find  an  empirical  equation  connecting  c  and  t  in  the  form  c  =  ab\ 

14.  The  relation  between  the  length  I  (in  millimeters)  and  the 
time  t  (in  seconds)  of  a  swinging  pendulum  is  found  as  follows : 


I          63.4 

80.5 

90.4 

101.3 

107.3 

140.6 

t         0.806 

0.892 

0.960 

1.010 

1.038 

1.198 

Find  an  empirical  equation  connecting  I  and  t  in  the  form  t  =  kl^. 

15.  For  a  dynamometer  the  relation  between  the  deflection  6, 

27r 
when  the  unit  0  =  -r— >  and  the  current  /,  measured  in  amperes,  is 

as  follows : 


e 

40 

86 

120 

160 

201 

240 

280 

320 

362 

I 

0.147 

0.215 

0.252 

0.293 

0.329 

0.360 

0.390 

0.417 

0.442 

Find  an  empirical  equation  connecting  /  and  0  in  the  form  /  =  kG^. 

16.  In  a  chemical  experiment  the  relation  between  the  concen- 
tration y  of  undissociated  hydrochloric  acid  and  the  concentration  x 
of  hydrogen  ions  is  shown  in  the  table : 


X 

1.68 

1.22 

0.784 

0.426 

0.092 

0.047 

0.0096 

0.0049 

y 

1.32 

0.676 

0.216 

0.074 

0.0085 

0.00315 

0.00036 

0.00014 

Find  an  empirical  equation  connecting  the  two  quantities  in  the 
form  y  =  kx". 


k 


170    EXPONENTIAL  AND  LOGARITHMIC  FUNCTIONS 

17.  Assuming  Boyle's  law,  ^z;  =  c,  determine  c  graphically  from 
the  following  pairs  of  observed  values : 


p 

39.92 

42.17 

45.80 

48.52 

51.89 

60.47 

65.97 

V 

40.37 

38.32 

35.32 

33.29 

31.22 

26.86 

24.53 

Find  -~  in  each  of  the  following  cases : 

3x-2 


19.  y  =  In  sinar. 

20.  y  =  tan-i — 


21.  2/  =  In  (2a^  +  V4  a;2  -  1)  +  2  x  csc-i2  a;.         / 

_2 

22.  y  =  x^e   * 

23.  y  =  ln^^^^-e-H 

24.  y  =  ^  tan^  aa;  +  In  cos  aa;. 
26.  y  =  a;  tan-^a;  —  ^  In (1  +  a;^). 

26.  A  substance  of  amount  x  is  being  decomposed  at  a  rate  which 
is  proportional  to  x.  If  a;  =  3.12  when  ^  =  0,  and  x  =  1.36  when 
^  =  40  min.,  find  the  value  of  x  when  ^  =  1  hr. 

27.  A  substance  is  being  transformed  into  another  at  a  rate  which 
is  proportional  to  the  amount  of  the  substance  still  untransformed. 
If  the  amount  is  50  when  ^  =  0,  and  15.6  when  ^  =  4  hr.,  find  how 
long  it  will  be  before  ^^^  of  the  original  substance  will  remain. 

28.  According  to  Newton's  law  the  rate  at  which  the  temperature 
of  a  body  cools  in  air  is  proportional  to  the  difference  between  the 
temperature  of  the  body  and  that  of  the  air.  If  the  temperature  of 
the  air  is  kept  at  60°,  and  the  body  cools  from  130°  to  120°  in  300  sec, 
when  will  its  temperature  be  100°  ? 

29.  Assuming  that  the  rate  of  change  of  atmospheric  pressure  p 
at  a  distance  h  above  the  surface  of  the  earth  is  proportional  to  the 
pressure,  and  that  the  pressure  at  sea  level  is  14.7  lb.  per  square  inch 
and  at  a  distance  of  1600  ft.  above  sea  level  is  13.8  lb.  per  square 
inch,  find  the  law  connecting  p  and  // . 


GENERAL  EXERCISES  171 

30.  Prove  that  the  curve  y  =  e~^^  sin  3  ic  is  tangent  to  the  curve 
y  =  e-2a:  at  any  point  common  to  the  two  curves. 

31.  At  any  time  t  the  coordinates  of  a  point  moving  in  a  plane 
are  x  =  e~^'  cos  2  t,  y  =  e~'^^  sin  2  t.  Find  the  velocity  of  the  point  at 
any  time  t.  Find  the  rate  at  which  the  distance  of  the  point  from 
the  origin  is  decreasing.  Prove  that  the  path  of  the  point  is  a  loga- 
rithmic spiral. 

32.  Show  that  the  logarithmic  spiral  r  =  e"^  cuts  all  radius  vectors 
at  a  constant  angle. 

33.  Find  the  radius  of  curvature  of  the  curve  y  =  e~^''  sin  2cc  at 
the  point  for  which  x  —  —- 

34.  Show  that  the  catenary  y  = -\e'* -\- e  "/   and  the  parabola 

1  "^ 

y  =  a  -\-  ^—  x"^  have  the  same  slope  and  the  same  curvature  at  their 
Z  a 

common  point. 

35.  Find  the  radius  of  curvature  of  the  curve  x  —  e^  sin  t,y  =  e^  cos  t. 

36.  Find   the    radius    of   curvature    of   the    curve   x  =  a  cos^  ^, 

77" 

y  —  a  sin^  <^,  at  the  point  for  which  <^  =  j-- 

37.  Find  the  radius  of  curvature  of  the  curve  y  =  \nx  and  its 
least  value. 

38.  Find  the  radius  of  curvature  of  the  curve  y  —  e^  cos  a;  at  the 
point  for  which  cc  =  —  • 


CHAPTER  YII 


SERIES 


58.  Power  series.    The  expression 


%  +  ct^x  -\r  a^r  +  a^x^  +  a^^  + 


(1) 


where  a^^  a^ 


a^,  •  •  •  are  constants,  is  called  a  power  series  in  x. 
The  terms  of  the  series  may  be  unlimited  in  number,  in  which 
case  we  have  an  infinite  series,  or  the  series  may  terminate  after 
a  finite  number  of  terms,  in  which  case  it  reduces  to  a  polynomial. 

If  the  series  (1)  is  an  infinite  series,  it  is  said  to  converge 
for  a  definite  value  of  x  when  the  sum  of  the  first  n  terms 
approaches  a  limit  as  n  increases  indefinitely. 

Infinite  series  may  arise  through  the  use  of  elementary  opera- 
tions. Thus,  if  we  divide  1  by  1  —  a;  in  the  ordinary  manner, 
we  obtain  the  quotient 

l+x  +  x'+x^-] , 

and  we  may  write 

1         .........  ^2) 


1  +  x  +  x^+x^+x^^- 


Similarly,  if  we  extract  the  square  root  of  1  +  a^  by  the  rule 
taught  in  elementary  algebra,  arranging  the  work  as  follows: 


l-\-x 
1 


1  +  £_^_|. 
2      8 


^-i 

X 
X 

4 

2  +  x 

x' 
8 

x' 
4 

4 

x^       x^ 
8  "^64' 

172 


POWER  SERIES  173 


the  operation  may  be  continued  indefinitely.    We  may  write 

-2 


I 


X         X 


'     ViT^  =  l  +  |-|+---  .      (3) 

The  results  (2)  and  (3)  are  useful  only  for  values  of  x  for 
which  the  series  in  each  case  converges.  When  that  happens 
the  more  terms  we  take  of  the  series,  the  more  nearly  is  their 
sum  equal  to  the  function  on  the  left  of  the  equation,  and  in 
that  sense  the  function  is  equal  to  the  series.  For  example,  the 
series  (2)  is  a  geometric  progression  which  is  known  to  con- 
verge when  ic  is  a  positive  or  negative  number  numerically  less 
than  1.    If  we  place  x  —  \\xi  (2),  we  have 

3— 14-1-|_1_|_     1     -L     1     J_... 

which  is  true  in  the  sense  that  the  limit  of  the  sum  of  the  terms 
on  the  right  is  |.    If,  however,  we  place  rr  =  3  in  (2),  we  have 

-i-=l-|-3  +  9  +  27+..., 

which  is  false.    A  reason  for  this   difference  may  be  seen  by 

considering  the  remainder  in  the  division  which  produced  (2) 

but  which  is  neglected  in  writing  the  series.    This  remainder  is 

a:" 

after  n  terms  of  the  quotient  have  been  obtained ;  and  if 

\—x 

X  is  numerically  less  than  1,  the  remainder  becomes  smaller  and 
smaller  as  n  increases,  while  if  x  is  numerically  greater  than  1, 
the  remainder  becomes  larger.  Hence  in  the  former  case  it  may 
be  neglected,  but  not  in  the  latter  case. 

The  calculus  offers  a  general  method  for  finding  such  series 
as  those  obtained  by  the  special  methods  which  led  to  (2)  and 
(3).    This  method  will  be  given  in  the  following  section. 

59.  Maclaurin's  series.  We  shall  assume  that  a  function  can 
usually  be  expressed  by  a  power  series  which  is  valid  for  ap- 
propriate values  of  x^  and  that  the  derivative  of  the  function  may 
be  found  by  differentiating  the  series  term  by  term.  The  proof 
of  these  assumptions  lies  outside  the  scope  of  this  book.  Let  us 
proceed  to  find  the  expansion  of  sin  x  into  a  series.  We  begin  by 
writing       &mx=A  +  Bx  +  Cx''-{-Dx''+Ex'  +  Fx''^-  '  '  ',  (1) 

where  A^  B,  C,  etc.  are  coefficients  to  be  determined. 


174  SERIES 

By  differentiating  (1)  successively,  we  have 

cosx  =  B-\-2Cx  +  SDx''+^  Ex^-{-  5  Fx'-\-  •  .  ., 

-  sin  X  =  2  C  +  3  .  2  .  D^  +  4  .  3  .  ^V+  5  .  4  .  Fx^'^  •  •  ., 

-  cos  a;  =  3  .  2  .  i)  +  4  .  3  .  2  .  ^o:  4-  5  .  4  .  3i<V  +  .  .  ., 

sin  a:  =  4  .  3  .  2  .  ^^  +  5  .  4  .  3  .  2  .  i^^^  H , 

cosa:=5.4  -3  '2F+  .  •  .. 

By  substituting  a:  =  0  in  equation  (1)  and  each  of  the  fol- 
lowing equations,  we  get 

^=0,   B  =  l,    C=0,   3.2.i)  =  -l,  F=0,   5  .4  .3  .2.i<'  =  l; 

whence  ^  =  0,     ^  =  1,      (7=0,     D  =  ~-—,      F=0,     F=^- 

o  !  5 ! 

Substituting  these  values  in  (1),  we  have 

sina;  =  :c-|^-f  ^ ,  (2) 

and  the  law  of  the  following  terms  is  evident. 

The  above  method  may  obviously  be  used  for  any  function 
which  may  be  expanded  into  a  series.  We  may  also  obtain  a 
general  formula  by  repeating  the  above  operations  for  a  general 
function /(a;). 

We  ^\dicef(x^  =  A  +  Bx  +  Cx''+Dx^-hFx'+  ...  (3) 

and,  by  differentiation,  obtain  in  succession 

f(x')  =  B-h2Cx-{-^Dx^-j-4:  Ex^+  .  .  ., 
/'(a:)  =  2!    C+3  .21)2: +  4.3^2:'+ ..., 
/^^(2:)  =  3!  i)  +  4.3.2.^a;+..., 
f-(x)  =  V.   ^+-.., 

where  /'(^),  /''(a:),  f"  (x),  and/^^  Qc)  represent  the  first,  second, 
third,  and  fourth  derivatives  oif(x). 

We  now  place  a:  =  0  in  these  equations,  indicating  the  results 
of  that  substitution  on  the  left  of  the  equations  by  the  symbols 


MACLAURIN'S  SERIES  175 

/(0),/(0),  /"(O),  etc.  We  thus  determine  A,  B,  C,  D,  E,  etc., 
and,  substituting  in  (3),  have 

/(a.)=/(0)+/'(0):.+l/'(0):.^+i/"(0):.'+i/v(oy+....(4) 

This  is  called  Maclaurin^s  series, 

Ex.  1.  Find  the  value  of  sin  10°  to  four  decimal  places. 
We  may  use  series  (2),  but  have  to  remember  (§  42)  that  x  must  be  in 
circular  measure.    Hence  we  place  x  = =  .17453,  where  we  take  five 

significant  figures  in  order  to  insure  accuracy  in  the  fourth  significant 
figure  of  the  result.* 

Substituting  in  (2),  we  have 

.     TT        ,_._„      (.17453)3 
sm  -—  =  .17453  —  ^^ — - — ^  +  •  •  • 

18  6 

=  .17453  -  .00089  =  .17364. 

Hence  to  four  decimal  places  sin  18°  =  .1736. 

We  have  used  only  two  terms  of  the  series,  since  a  rough  calculation, 
which  may  be  made  with  x  —  .2,  shows  that  the  third  term  of  the  series 
will  not  affect  the  fourth  decimal  place. 

Ex.  2.  Find  the  value  of  sin  61°  to  four  decimal  places. 

In  radians  the  angle  61°  is  - — rr  =  1.0647.  If  this  number  were  sub- 
stituted in  the  series  (2),  a  great  many  terms  would  have  to  be  taken  to 
include  all  which  affect  the  first  four  decimal  places.    We  shall  therefore 

find  a  series  for  sin(-+a;)  and  afterwards   place    a:  =  — —  (=1°).    We 
^       \3         /  180 

chose  the  angle  -(=  60°)  because  it  is  an  angle  near  61°  for  which  we 

o 

know  the  sine  and  cosine.  The  series  may  be  obtained  by  the  method  by 
which  (2)  was  obtained.  For  variety  we  shall  use  the  general  formula  (4). 
We  have  then  v  1 

fix)  =  sing  +  rj,      /(O)  =  sin^  =  -V3, 


b 


/(:r)  =  cosg  +  x),     /(O)  =  cos|  =  i, 
f\x)=-  sin/|  +  a:),    /X0)  =  -  sin?  =  -  i V3. 


3  2 


*  This  is  not  a  general  rule.  In  other  cases  the  student  may  need  to  carry 
two  or  even  three  more  significant  figures  in  the  calculation  than  are  needed 
in  the  result. 


176  SERIES 

Therefore,  substituting  in  (4),  we  have 

In  this  we  place  x  —  — —  =  .01745  and  perform  the  arithmetical  calcula- 
/  \ 

tion.    We  have  sin  61°  =  sin  (J  +  r|- )  =  .8746. 

\o      180/ 

Ex.  3.    Expand  In  (1  ^  x). 

The  function  In  x  is  an  example  of  a  function  which  cannot  be  expanded 
into  a  Maclaurin's  series,  since  if  we  place  f{x)  =  In  a:,  we  find  /(0),/'(0), 
etc.  to  be  infinite,  and  the  series  (4)  cannot  be  written.  We  can,  however, 
expand  In  (1  +  x)  by  series  (4)  or  by  using  the  method  employed  in  obtain- 
ing (2).  The  latter  method  is  more  instructive  because  of  an  interesting 
abbreviation  of  the  work.    We  place 

\ti(1  +  x)  =  A  +  Bx+  Cx^  +  Dx^  +  Ex^-{-  "'. 

Then,  by  differentiating, 

— i—  =  B  +  2  Cx  +  d  Dx^  -{-  ^  Ex^  +  "  : 
1  +  X 

But  we  know,  by  elementary  algebra,  that 

1  —  X  +  x^  —  x^  -\-  •  •  • . 


1  +  a; 
Hence,  by  comparing  the  last  two  series,  we  have 

B  =  l,     C  =  -i,     D  =  ^,     E=-\,     etc. 

By  placing  x  =  0  in  the  first  series,  we  find  In  1  =  ^,  whence  ^  =  0.  We 
have,  therefore,  2      ^3      ^4 

ln(l  +  x)  =  x-|  +  |-|+.... 

EXERCISES 

Expand  each  of  the  following  functions  into  a  Maclaurin's  series  : 
1.  e^.  2.  cosic.  3.  tana;.  4.  sin~^a:;. 

5.  tan-^ic.  6.  sinfj-l-ic].  7.  \n(2-\-x). 

8.  Prove  the  binomial  theorem 

iW  I  o  I 

9.  Compute  sin  5°  to  four  decimal  places. 
10.  Compute  cos  62°  to  four  decimal  places. 


TAYLOR'S  SERIES  177 

60.  Taylor's  series.  In  the  use  of  Maclaurin's  series,  as  given 
in  the  previous  section,  it  is  usually  necessary  to  restrict  our- 
selves to  small  values  of  x.  This  is  for  two  reasons.  In  the 
first  place,  the  series  may  not  converge  for  large  values  of  x\ 
and  in  the  second  place,  even  if  it  converges,  the  number  of 
terms  of  the  series  which  it  is  necessary  to  take  to  obtain  a 
required  degree  of  accuracy  may  be  inconveniently  large.  This 
difficulty  may  be  overcome  by  an  ingenious  use  of  Maclaurin's 
series  as  illustrated  in  Ex.  2  of  the  previous  section.  We  may, 
however,  obtain  another  form  of  series  which  may  be  used  when 
Maclaurin's  series  is  inconvenient. 

Let  f(x)  be  a  given  function,  and  let  a  be  a  fixed  value  of 
X  for  which  the  values  of  f(x)  and  its  derivatives  are  known. 
Let  ic  be  a  variable,  or  general,  value  of  x  which  does  not  differ 
much  from  a ;  that  is,  let  a:  —  a  be  a  small  number,  positive  or 
negative.  We  shall  then  assume  that /(a:)  can  be  expanded  in 
powers  of  the  binomial  x  —  a-^  that  is,  we  write 

f(x)=A-\-B(x-a)  +  C(x-ay  +  D(x-ay+^.^,     (1) 

and  the  problem  is  to  determine  the  coefficients  A^  B^  C,  •  •  •. 
We  differentiate  equation  (1)  successively,  obtaining 

f(x)  =  B-ir'lC(x-a)-\-ZD(x-ay-ir  •  •  •, 
/^(a:)=  2C+ 3;2D(a^- a)4- •  •  ., 

In  each  of  these  equations  place  x  =  a.    We  have 
f(a)=A,    f(a)=B,    f"(a)=2}lC,     etc.; 

whence    ^  =/(«),     B  =f  (a),     ^=*^yP'     ^^^~iP''     ^^' 

Substituting  in  equation  (1),  we  have  as  the  final  result 

f(^^y=f(a)+fCaXx-a)+-^Cx-ay+-^Cx-ay+  ■ ...  (2) 

This  is  known  as  Taylor^ s  series.  Since,  as  has  been  said,  it 
is  valid  for  values  of  x  which  make  a:  —  a  a  small  quantity,  the 


178  SERIES 

function  f(x)  is  said  to  be  expanded  in  the  neighborhood  of 
x  =  a.  It  is  to  be  noticed  that  Taylor's  series  reduces  to 
Maclaurin's  series  when  a  =  0.  Maclaurin's  series  is  therefore 
an  expansion  in  the  neighborhood  of  a;  =  0. 

Ex.    Expand  In  a:  in  the  neighborhood  of  x  =  3. 

Here  we  have  to  place  a  =  3  in  the  general  formula.  The  calculation  of 
the  coefficients  is  as  follows  : 

f{x)  =  hix,       /(3)  =  ln3, 

/'(^)=^-^2'  r(3)=-i 

and  therefore 

In  a;  =  In  3  +  i  (a;  -  3)  -  ^^{x  -  3)2  +  ^^  {x  -  3)^  +  •  •  •. 

This  enables  us  to  calculate  the  natural  logarithm  of  a  number  near  3, 
provided  we  know  the  logarithm  of  3.  For  example,  let  us  have  given 
In  3  =1.0986  and  desire  In  3^.    Then  x  —  3  =  ^,  and  the  series  gives 

ln3i=1.0986  +  ^-7V  +  a^  +  --- 

=  1.0986  +  .1667  -  .0139  +  .0015  +  •  •  • 
=  1.2529. 

The  last  figure  cannot  be  depended  upon,  since  we  have  used  only 
four  decimal  places  in  the  calculation. 

EXERCISES 

Expand  each  of  the  following  functions  into  a  Taylor's  series, 
using  the  value  of  a  given  in  each  case  : 

'  4.  cos  X.    a  =  —' 

1  " 

2. ,    a  =  2.  6-  6%    a  =  3. 

1  -f-  X 

73-  6.  tan~^a:;,    a-  =  1. 

3.  since,    0^  = -T-  „       /7— — 5  h 

'4  7.  ■\/l-\-x\    a  —  1. 


8.  Compute  sin  46°  to  four  decimal  places  by  Taylor's  series. 

9.  Compute  cos  32°  to  four  decimal  places  by  Taylor's  series. 
10.  Compute  e^*^  to  four  decimal  places  by  Taylor's  series. 


GENERAL  EXERCISES  179 

GENERAL  EXERCISES 

Expand  each  of  the  following  functions  into  series  in  powers  of  x : 

^  ^  4.  In  3 6.  sin(  — +  ic)- 

2.  secx.  ^  —  x  \6         / 

1  ^  ('^        ^  -  ^ 

5.  cos 


(f-> 


Vl+  x"  V3         /  -Wl-x" 

8.  Verify  the  expansion  of  tana; (Ex.  3,  §59)  by  dividing  the 
series  for  sin  x  by  that  for  cos  x. 

9.  Verify  the  expansion  of  sec  a;  (Ex.  2)  by  dividing  1  by  the 

series  for  cos  x. 

1—  X 

10.  Expand  by  Maclaurin's  series  and  verify  by  dividing 

the  numerator  by  the  denominator. 

11.  Expand  e^cos  a?  into  a  Maclaurin's  series,  and  verify  by 
multiplying  the  series   for  e^  by  that  for  cos  x. 

12.  Expand  e^sina;  into  a  Maclaurin's  series,  and  verify  by 
multiplying  the  series  for  e^  by  that  for  since. 

13.  Expand  e^ln(l+a;)  into  a  Maclaurin's  series,  and  verify  by 
multiplying  the  series  for  e^  by  that  for  In  (1  -|-  a^). 

14.  Compute  cos  15°  to  four  decimal  places. 

15.  Compute  sin  31°  to  four  decimal  places. 

16.  Compute  e*  to  four  decimal  places  by  the  series  found  in 
Ex.  1,  §  59. 

17.  Using  the  series  for  ln(l4-a:;),  compute  In  |  to  five  decimal 
places. 

18.  Using  the  series  found  in  Ex.  4,  compute  In  2  to  five  decimal 
places,  and  thence,  by  aid  of  the  result  of  Ex.  17,  find  In  3  to  four 
decimal  places. 

19.  Using  the  series  found  in  Ex.  4,  compute  In  |  to  five  decimal 
places,  and  thence,  by  aid  of  the  first  result  of  Ex.  18,  find  In  5  to 
four  decimal  places. 

20.  Using  the  series  found  in  Ex.  4,  compute  In  ^  to  four  decimal 
places,  and  thence,  by  aid  of  the  result  of  Ex.  18,  find  In  7  to  three 
decimal  places. 

21.  Compute  the  value  of  tt  to  four  decimal  places,  from  the  ex- 

1         TT 

pansion  of  sin~^a;  (Ex.  4,  §  59)  and  the  relation  sin~^-  =  -  • 


180  SERIES 

22.  Compute  the  value  of  tt  to  four  decimal  places,  from  the  ex- 

1  1         TT 

pansion  of  tan~^x  (Ex.  5,  §  59)  and  the  relation  tan~^-  +  2  tan~^-  =  —  • 

23.  Compute  •\/ll  to  four  decimal  places  by  the  binomial  theorem 
(Ex.  8,  §  59),  placing  a  =16,  x  ==  1. 

24.  Compute  "^26  to  four  decimal  places  by  the  binomial  theorem 
(Ex.  8,  §  59),  placing  a  =  27,  a;  =- 1. 

'     dx   in   the    form    of    a    series 

26.  Obtain    the    integral    /    e'^^dx    in    the    form    of    a    series 
expansion.  ^^ 

' in  the  form  of  a  series  expansion. 

JC^    dx     . 
I    ij ^  in  the  form  of  a  series  expansion. 


CHAPTER  Vni 

PARTIAL  DIFFERENTIATION 

61.  Partial  differentiation.  A  quantity  is  a  function  of  two 
variables  x  and  y  when  the  values  of  x  and  y  determine  the 
quantity.  Such  a  function  is  represented  by  the  symbol /(a;,  y). 
For  example,  the  volume  F  of  a  right  circular  cylinder  is  a 
function  of  its  radius  r  and  its  altitude  A,  and  in  this  case 

Similarly,  we  may  have  a  function  of  three  or  more  variables 
represented  by  the  symbols /(a:,  y,  z^,f(x,  y,  z,  u),  etc. 

Consider  now  /(a;,  ^),  where  x  and  y  are  independent  varia- 
bles so  that  the  value  of  x  depends  in  no  way  upon  the  value 
of  y  nor  does  the  value  of  y  depend  upon  that  of  x.  Then  we 
may  change  x  without  changing  y^  and  the  change  in  x  causes 
a  change  in  /.  The  limit  of  the  ratio  of  these  changes  is  the 
derivative  of  /  with  respect  to  x  when  y  is  constant,  and  may 

be  represented  by  the  symbol /-r^  J  • 

Similarly,  the  derivative  of  /  with  respect  to  y  when  x  is 

constant,  is  represented  by  the  symbol  l^\    These  derivatives 

are  called  partial  derivatives  of  /  with  respect  to  x  and  y  re- 
spectively. The  symbol  used  indicates  by  the  letter  outside 
the  parenthesis  the  variable  held  constant  in  the  differentiation. 
When  no  ambiguity  can  arise  as  to  this  variable,  the  partial  de- 

rivatives  are  represented  by  the  symbols  ^  and  --,  thus: 

^/^m  _^._fix  +  ^x,  y^-f(x,y) 
\dx), 


I  ,    ;       Lim 

ex      \ax/„     Ax-o  Ax 


dy      \dylx     Ay-^o  Ay 

181 


182  PARTIAL  DIFFERENTIATION 

So,  in  general,  if  we  have  a  function  of  any  number  of  variables 
f(x^  ?/,...,  2),  we  may  have  a  partial  derivative  with  respect  to 
each  of  the  variables.   These  derivatives  are  expressed  by  the  sym- 

To  compute  thfese  derivatives  we  have  to  apply  the  formulas 
for  the  derivative  of  a  function  of  one  variable,  regarding  as 
constant  all  the  variables  except  the  one  with  respect  to  which 
we  differentiate. 

ct 
Ex.  1.  Consider  a  perfect  gas  obeying  the  law  v  =  —  •    We  may  change 

P 

the  temperature  while  keeping  the  pressure  unchanged.    If  At  and  At;  are 

corresponding  increments  of  t  and  v,  then 

.     _  c(t  +  At)       ct  _  cAt 

and 


p  P        P 

dv  _  c 
dt~p' 


Or  we  may  change  the  pressure  while  keeping  the  temperature  un- 
changed.   If  Ap  and  Ay  are  corresponding  increments  of  p  and  v,  then 

ctAp 


Av-       '^     - 

ct 

p  +  Ap 

P 

-                                     dv           ct 

and                                  —  = . 

dp         p^ 

■     Ex,2.  /=a:3-3a:2?/  +  ?/3. 

1  =  3.-6.,, 

2f  =  -3x2  +  32^2. 

^2  +^AP 


Ex.  3.  f=Bui{x'^  +  y'^), 

2  X  cos  (.2  +  3/2-j, 


y  -  o  ^  .^o  r^2    ,    „2> 


dx 

^  =  2  3/  cos  (.2  +  y^). 

vy  dy 

Ex.  4.    In  differentiating  in  this  way  care  must  be  taken  to  have  the 
functions  expressed  in  terms  of  the  independent  variables.    Let 

X  =  r  cos  6y  y  =  r  sin  0- 

Then  —  =  cos  6,  —  =  sin  6, 

dr  dr 

dx    ...         .     /,  dy  ^ 

where  r  and  6  are  the  independent  variables. 


(1) 


PARTIAL  DERIVATIVES 


183 


Moreover,  since  r  =  Var^-j.  y'^  and  6  =  tan~^ -> 


=  cos  6, 

_      sin  5 
r 


y 


dr_ 

S5  _       a:       _  cos  ^ 


sin^, 


(2) 


where  x  and  y  are  the  independent  variables. 

dx 


dr  . 


It  is  to  be  emphasized  that  —  in  (1)  is  not  the  reciprocal  of  —  in  (2). 
InfacMn(l),|=g)^anj:(.),|=@. 

and  because  the  variable  held  constant  is  differ- 
ent in  the  two  cases,  there  is  no  reason  that 
one  should  be  the  reciprocal  of  the  other.  It 
happens  in  this  case  that  the  two  are  equal,  but 
this  is  not  a  general  rule.  Graphically  (Fig.  80), 
if  OP  =  r  is  increased  by  PQ  =  Ar,  while  6  is 
constant,  then  PR  =  Ax  is  determined.  Then 
dx  _  /dx\ 
dr       \drj e 

Moreover  (Fig.  81),  if  OM  =  a:  is  increased  by 
MN=  PQ  =  Aa:,  while  y  is  constant,  then  RQ  =  Ar 

RQ  ^ 

PQ 

dx 


J.     PR  ^ 

Lim  — —  =  cos  a. 
PQ 


determined.     Then   —  =  (  — )  =  Lim 
dx      \dxly 


cos  0.    It  happens  here  that  —  =  —    But 

dd  ^''     ^-^ 

in  (1),  and  — ,  in  (2),  are  neither  equal  nor 

.^  ^      ,       aa;  ^  ^ 

reciprocal. 

EXERCISES 

Find  77-  and  75-  in  each  of  the  following  cases : 
ex  cy 


1.  z==  x^—4:X^y  +  ^xy'^-\-by^. 


5.  z  =  \n 


xy 


2.  z 


_     ^y 


3.  z  =  ctn 


4.  z  =  sm~'^xy. 


6.  z 


7.  z 


sin 


x  —  y 
2xy 


x-\-y 


S.  z  =  ln(a;  -f-  Vx^  -f  f). 


9.  If  ;^  =  ln(^2  _  2xy  ^^-\-Sx  -Sy),  prove  -^ -^  —  =  0. 

/ ^  dz         dz 

10.  If  ^  =  Wx^  +  i/^e-^,  prove  x-^  -\-  y-^  —  z. 


184  PARTIAL  DIFFERENTIATION 

62.  Higher  partial  derivatives.  The  partial  derivatives  of 
f(x^  y)  are  themselves  functions  of  x  and  y  which  may  have 
partial  derivatives,  called  the  second  partial  derivatives  oif(x,  ?/). 

T'-^y  '"^  £(!)'  4©'  £©'  4(1)'  ^"* ''  "'^  ^' ''°"" 

that  the  order  of  differentiation  with  respect  to  x  and  y  is  imma- 
terial when  the  functions  and  their  derivative  fulfill  the  ordinary- 
conditions  as  to  continuity,  so  that  the  second  partial  derivatives 
are  three  in  number,  expressed  by  the  symbols 


dx 


\dx)    z^   -^^ 


\dy)      dy\dx)      =~«-     •^''" 


dx\dy/      dy\dx/     8xdy 
dy\dy)      df     ■'"• 


Similarly,  the  third  partial  derivatives  of  f(Xj  y')   are  four  in 
number;  namely. 


dxKdx")      ex''' 


dy\dx^)      dx\dxdyl      d3?\dy/       da^dy 

d  m\_  d  /  dy\     c'  /df\     ay  ^ 

dx\dy^/      dy\dxdy/      dy^\dx/      dxdy^ 
dy\df)~dy'' 

^P  +  Qf 

So,  in  general,  - — ~  signifies  the  result  of  differentiating 

if 

f(xj  y)  p  times  with  respect  to  x,  and  q  times  with  respect  to 
y,  the  order  of  differentiating  being  immaterial. 

In  like  manner,      ^^  signifies  the  result  of  differentiating 

if 

f(x,  y,  z)  p  times  with  respect  to  x,  q  times  with  respect  to  y, 
and  r  times  with  respect  to  2,  in  any  order. 


TOTAL  DIFFERENTIAL  186 


EXERCISES 


1.  Uz  =  (x^-{-  if)  tan-i^,  find    ^'^ 


2.  If  z  =  eP  sin  (a;  —  y),  find  ^• 

3.  If;.  =  ln(a;2+2/^),  find^. 


df 
Verify  Yy^]~Y\^]  ^"^  ^^^  ^^  *^^  following  cases 


1  X 


4.  z  =  xy'^-\-2ye''.  6.  ^  =  sin 

5.  .  =  i^^+i^.  7..=         * 


8.  If  ;^  =  tan-i-,  prove  ^  +  ^-^  =  0. 

9.  If  ^  =  In  {x"  -  ahf),  prove  «^'  ^  -  ^2  =  0- 

^2  (7.7)  d'^V 

10.  If  F  =  r"'  cos  n<^y  prove  wV  — )  ^^  -{■  7n{m  -\-  1)^t^  =  0. 

63.  Total  differential  of  a  function  of  two  variables.    In  §  20 

the  differential  of  a  function  of  a  single  variable,  y  =fQjc)^  is 
defined  by  the  equation     ay=f(x)dx,  (1) 

where  f  {x)  is  the  derivative  of  y. 

But  /(rr)=Lim(^);  (2) 

and  hence,  according  to  the  definition  of  a  limit  (§  1), 

g=/(.)  +  e.  (3) 

where  €  denotes  the  difference  between  the  variable  -^  and  its 

limit /'(a;)  and  approaches  zero  as  a  limit  as  Lx—^  0. 
Multiplying  (3)  by  Aa:,  we  have 

A^  =/  {x)  A2:  +  e  b.x.  (4) 

But  A3:  =  dx  and  A?/  =f{x  +  A2;)  -f(x),  so  that  (4)  may  be 
written  in  the  form 

f{x  +  Aa:)  -/(^)  =/  (2-)  ^2:  +  €  dx.  (5) 


186  PARTIAL  DIFFERENTIATION 

In  the  case  of  a  function  of  two  variables, /(a;,  y)^iix  alone 
is  changed,  we  have,  by  (5), 

f(x  +  A2:,  y)  -f(x,  y)  =  -£dx^-  e^dx,  (6) 

the  theory  being  the  same  as  in  the  case  of  a  function  of  one 

variable,  since  y  is  held  constant.  The  term  -^  dx  may  be  denoted 
by  the  symbol  d^f. 

Similarly,  if  x  is  held  constant  and  y  alone  is  changed,  we  have 

f(x,  y  +  Ay) -fix,  y)JI^dy  +  e^dy,  (7) 

and  -^dy  may  be  denoted  by  the  symbol  dj^f. 

Finally,  let  x  and  y  both  change.    Then 
A/=/(:?^+ A2:,  2/  + A^)  -f{x,  y) 

=/(ic+ Aa;,  y  +  Ay)-f(xi-Ax,  y)+f(x  +  Ax,y^-f(x,y^.   (8) 

Then,  by  (6), 

f(x  +  Ax,  y) -f(x,  y')  =  £dx  +  e^dx ;  (9) 

and  similarly,  by  (7), 

f(x  +  Ax,  y  +  Ay)  -f(x  +  Ax,  y)  =  ^dy  +  edy,      (10) 

df  ^ 

where  ~  is  to  be  computed  for  the  value  (x-{-Ax,  y').    But  if 

df .     y    . 

r^  is  a  continuous  function,  as  we  shall  assume  it  is,  its  value 
dy 

for  (x  +  Ax,  y)  differs  from  its  value  for  (x,  «/)  by  an  amount 

which  approaches  zero  as  dx  approaches  zero.    Hence  we  may 

write,  from  (8),  (9),  and  (10), 

Af=%dx  +  ^^dy-^e^dx  +  e^dy,  (11) 

ex  cy 

where  both  -^  ^^^  ^  ^^^  computed  for  {x,  y). 

We  now  write  df=  —  dx+  —  dy,  (12) 

-^      dx  dy    ^'  ^     ^ 

so  that  Af=df+  e^dx  +  €^dy,  (1 3) 

and  df  is  called  the  total  differential  of  the  function,  the  expres- 
sions d^f  and  dj^  being  called  the  partial  differentials. 


TOTAL  DIFFERENTIAL  187 

It  is  evident,  by  analogy  with  the  case  of  a  function  of  a 
single  variable,  that  a  partial  differential  expresses  approximately 
the  change  in  the  function  caused  by  a  change  in  one  of  the 
independent  variables,  and  that  the  total  differential  expresses 
approximately  the  change  in  the  function  caused  by  changes  in 
both  the  independent  variables.  It  is  evident  from  the  defini- 
ti<»^*hat  df=.dj+dj.  (14) 

Ex.   The  period  of  a  simple  pendulum  with  small  oscillations  is 

>// 

4  ir'H 
whence  g  =  -—-  • 

Let  I  =  100  cm.  with  a  possible  error  of  ^  mm.  in  measuring,  and 
T  =  2  sec.  with  a  possible  error  of  ^^^  sec.  in  measuring.    Then  dl  =  ±  i^ 

andrfr=±xk-  4^2  8^2; 

Moreover,  dg=—dl-  -yg-  dT, 

and  we  obtain  the  largest  possible  error  in  g  by  taking  dl  and  dT  oi  oppo- 
site signs  say  dl  =  5'^,  dT  =  —  ^^q. 

Then  dg  =  ^  +  Tr^  =  1.05  ir^  =  10.36. 

The  ratio  of  error  is 

'IR  =  €-2  —  =  .0005  +  .01  =  .0105  =  1.05%. 
g        I  T 

EXERCISES 

1.  Calculate  the  numerical  difference  between  A^  and  dz  when 
z  ~  ^xy  —  x^  —  y'^,  X  =  2,  y  =  ^,  ^.x  =  dx  =  .01,  and  A^/  =  dy  =.001. 

2.  An  angle  </>  is   determined  from  the  formula  <^  =  tan~^^  by 

X 

measuring  the  sides  x  and  3/  of  a  right  triangle.  If  x  and  y  are 
found  to  be  6  ft.  and  8  ft.  respectively,  with  a  possible  error  of  one 
tenth  of  an  inch  in  measuring  each,  find  approximately  the  greatest 
possible  error  in  <^. 

3.  If  C  is  the  strength  of  an  electric  current  due  to  an  electro- 
motive force  E  along  a  circuit  of  resistance  i?,  by  Ohm's  law 


188  PARTIAL  DIFFERENTIATION 

If  errors  of  1  per  cent  are  made  in  measuring  E  and  72,  find 
approximately  the  greatest  possible  percentage  of  error  in  com- 
puting C 

4.  If  F  denotes  the  focal  length  of  a  combination  of  two  lenses 
in  contact,  their  thickness  being  neglected,  and  f^  and  f^  denote  the 
respective  focal  lengths  of  the  lenses,  then 

1=1+1. 

F     A      f. 

If  /j  and  f^  are  said  to  be  6  in.  and  10  in.  respectively,  find  approx- 
imately the  greatest  possible  error  in  the  computation  of  F  from  the 
above  formula  if  errors  of  .01  in.  in  f^  and  0.1  in.  in/^  are  made. 

5.  The  eccentricity  e  of  an  ellipse  of  axes  2  a  and  2^  (a  >  Z»)  is 
given  by  the  formula  . 

e  = 

a 

The  axes  of  an  ellipse  are  said  to  be  10  ft.  and  6  ft.  respectively. 
Find  approximately  the  greatest  possible  error  in  the  determination 
of  e  if  there  are  possible  errors  of  .1  ft.  in  a  and  .01  ft.  in  h, 

6.  The  hypotenuse  and  one  side  of  a  right  triangle  are  respectively 
13  in.  and  5  in.  If  the  hypotenuse  is  increased  by  .01  in.,  and  the 
given  side  is  decreased  by  .01  in.,  find  approximately  the  change  in 
the  other  side,  the  triangle  being  kept  a  right  triangle. 

7.  The  horizontal  range  7t  of  a  bullet  having  an  initial  velocity  of 
v^,  fired  at  an  elevation  a,  is  given  by  the  formula 

v^  sin  2  cr 
R  = • 

Find  approximately  the  greatest  possible  error  in  the  computation 
of  R  if  v^  =  10,000  ft.  per  second  with  a  possible  error  of  10  ft.  per 
second,  and  a  =  60°  with  a  possible  error  of  1'  (take  g  =  32). 

8.  The  density  D  of  a  body  is  determined  by  the  formula 

w  —  w 

where  w  is  the  weight  of  the  body  in  air  and  ^/;'  the  weight  in  water. 
\i  w  =  244,000  gr.  and  w^  =  220,400  gr.,  find  approximately  the 
largest  possible  error  in  B  caused  by  an  error  of  5  gr.  in  w  and  an 
error  of  10  gr.  in  w'. 


RATE  OF  CHANGE  189 

64.  Rate  of  change.    The  partial  derivative  ^  gives  the  rate 

ex      • 

of  change  of  /  with  respect  to  x  when  x  alone  varies,  and  the 

partial  derivative  ^  gives  the  rate  of  change  of  /  with  respect 

to  1/  when  ?/  alone  varies.  It  is  sometimes  desirable  to  find  the 
rate  of  change  of  /  with  respect  to  some  other  variable,  t  Ob- 
viously, if  this  rate  is  to  have  any  meaning,  x  and  y  must  be 
functions  of  f,  thus  making  /  also  a  function  of  t    Now,  by  §  11, 

the  rate  of  change  of  /  with  respect  to  t  is  the  derivative  —. 

civ 

To  obtain  this  derivative  we  have  simply  to  divide  df,  as  given 
by  (12),  §  63,  by  dt,  obtaining  in  this  way 

<¥^^dxdf^dy 

dt      dx  dt      dy  dt '  ^  ^ 

The  same  result  may  be  obtained  by  dividing  A/,  as  given  by 
(11),  §  63,  by  A^  and  taking  the  limit  as  A^  approaches  zero  as 
a  limit. 

Ex.  1.  If  the  radius  of  a  right  circular  cylinder  is  increasing  at  the  rate 
of  2  in.  per  second,  and  the  altitude  is  increasing  at  the  rate  of  3  in.  per 
second,  how  fast  is  the  volume  increasing  when  the  altitude  is  15  in.  and  the 
radius  5  in.  ? 

Let  V  be  the  volume,  r  the  radius,  and  h  the  altitude.    Then 
V=7rrVi. 

^  ^^'  dt  ~    dr  dt'^  dh   dt 

=  2  7rrh—  +  irr^—- 
dt  dt 

By  hypothesis,  —  =  2,  —  =  3,  r  =  5,  7i  =  15.  Therefore  — -=  375  tt  cu.  in. 
^     *^j  dt  dt  dt 

per  second. 

The  same  result  may  be  obtained  without  partial  differentiation  by  ex- 
pressing V  directly  in  terms  of  t.  For,  by  hypothesis,  r  =  5  +  2  f,  ^  =15  +  3  /, 
if  we  choose  t  =  0  when  r=  d  and  h  — 15.    Therefore 

F=(375  +  375  <  +  120  ^2+12  ^3)7r ; 

whence  —  =  (375  +  240  i  +  36  f")  it. 

dt       ^ 

When  /  =  0,  - —  =  375  ir  cu.  in.  per  second,  as  before. 

dt 


190 


PARTIAL  DIFFERENTIATION 


Ex.  2.    The  temperature  of  a  point  in  a  plane  is  given  by  the  formula 

1 

The  rate  of  change  of  the  temperature  in  a  direction  parallel  to  OX  is, 
accordingly,  ^  _ ^        2x 

dx  ~      (a;2  +  yy ' 

which  gives  the  limit  of  the  change  in  the  temperature  compared  with  a 
change  in  x  when  x  alone  varies. 

Similarly,  the  rate  of  change  of  m  in  a  direction  parallel  to  0  F  is 
du  _  2  y 

J^''  (x^+yY 

Suppose  now  we  wish  to  find  the  rate  of  change  of  the  temperature  in  a 
direction  which  makes  an  angle  a  with  OX.  From  Fig.  82,  if  Pi{x^,  y^)  is 
a  fixed  point,  and  P(x,  y)  a  moving  point  y. 

on  the  line  through  P^  making  an  angle  a 
with  OX, and  s  is  the  distance  P^P,  we  have 


whence 


and 


P,R-- 

-  X 

-  x^=  s  COS  a, 

RP  = 

^y 

-  y^  =  s  sin  a 

X  - 

=  ^1 

+  s  cos  a, 

y^ 

=  yi 

+  s  sin  a, 

dx 

—  =  cos  or, 

ds 

dy 

~  =  sm  c 
ds 

i? 


R 


Fig.  82 


dx 


dy 


Replacing  <  by  s  in  formula  (1),  and  substituting  the  values  of  —  and 


ds 


—  which  we  just  found,  we  have 
ds 

du       du  .   du    . 

— -  =  —  cos  a  -] sm  a 

ds       dx  dy 

_  _2  X  cos  a  +  2  y  sin  a 

~  (x'-^yy 

Formula  (1)  has  been  written  on  the  hypothesis  that  x  and  ?/ 
are  functions  of  t  only.  If  x  and  i/  are  functions  of  two  vari- 
ables, t  and  8,  and  (1)  is  derived  on  the  assumption  that  t 
alone  varies,  we  have  simply  to  use  the  notation  of  §  61  to  write 

at  once  m=^f(^\.^f(il\,  (2) 

\dtj.    dx\dt)7 dy\dt), 
which  may  also  be  written  as 


¥^^fdx_^dldy^ 
dt      dx  dt      dy  dt 


(3) 


GENERAL  EXERCISES  191 

EXERCISES 

1.  li  z  =  e  Xj  X  =  sin  t,  y  =  cos  t,  find  the  rate  of  change  of  z 
with  respect  to  t. 

\-\-  X 

2.  li  z  =  tan- ^3 >  x  =  sin  ^,  ?/  =  cos  t,  find  the  rate  of  change 

~  y  TT 

of  z  with  respect  to  t  when  ^  =  «■  * 

3.  If  F=  (e**^  —  e'"^)  cos  ay,  prove  that  F  and  its  derivatives  in 
any  direction  are  all  equal  to  zero  at  the  point  (  0,  ^ )  • 

4.  If  V= —  ?  find  the  rate  of  change  of   V  at  the  point 

(1,  1)  in  a  direction  making  an  angle  of  45°  with  OX. 

5.  If  the  electric  potential  V  at  any  point  of  a  plane  is  given  by 
the  formula  F=  In  Vic^+  y^,  find  the  rate  of  change  of  potential  at 
any  point :  (1)  in  a  direction  toward  the  origin  ;  (2)  in  a  direction  at 
right  angles  to  the  direction  toward  the  origin, 

6.  If  the  electric  potential  V  at  any  point  of  the  plane  is  given 

V(cc  —  af-^-iP- 
by  the  formula   F=  In  — ,  '  — ~  >  find  the  rate  of  change  of 

■>J{x^af^f   ^ 

potential  at  the  point  (0,  a)  in  the  direction  of  the  axis  of  y,  and  at 

the  point  {a,  a)  in  the  direction  toward  the  point  (—  a,  0). 

GENERAL  EXERCISES 

.    xy  —1  dz  dz 

1.  it  z  =  sm  -^ )  prove  x-:: ?/  ^  =  U. 

xy  -\-l    ^  dx      ^  cy 

1  '  dz  dz 

-  dz  dz 

3.  li  z  =  y^-{-  ye"",  prove  x^  -^  +  y  -^  =  ^  y^- 

d^z       d^z 

4.  If  ^  =  e-"^  cos  a(k  —  x),  prove  that  —  +  — ^  =  0. 

5.  liz  =  e-(^^+«='^-^)'  sin  kx,  prove  that  ^  =  a^  _J  _  ^2^^ 

6.  If    s  =  e-^^'sin  {my  +  a?  -Wa^m^—  k^),   prove    that   ^  +  2  A;  ^ 


=  ^2 


d?f 

d^V      1  ^F       1  d'^V 
7.  If  F=  e'^'^cos  (a  In  r),  prove  that  -^  +  -  ^  +  -^  ^  =  0. 


192  PARTIAL  DIFFERENTIATION 

8.  A  right  circular  cylinder  has  an  altitude  8  ft.  and  a  radius 
6  ft.  Find  approximately  the  change  in  the  volume  caused  by  de- 
creasing the  altitude  by  .1  ft.  and  the  radius  by  .01  ft. 

9.  The  velocity  v,  with  which  vibrations  travel  along  a  flexible 
string,  is  given  by  the  formula         r— 

where  t  is  the  tension  of  the  string  and  m  the  mass  of  a  unit  length 
of  it.  Find  approximately  the  greatest  possible  error  in  the  compu- 
tation of  V  if  ^  is  found  to  be  6,000,000  dynes  and  m  to  be  .005  gr. 
per  centimeter,  the  measurement  of  t  being  subject  to  a  possible 
error  of  1000  dynes  and  that  of  m  to  a  possible  error  of  .0005  gr. 

10.  The  base  AB  of  a  triangle  is  12  in.  long,  the  side  J.  C  is  10  in., 
and  the  angle  A  is  60°.  Calculate  the  change  in  the  area  caused 
by  increasing  ^  C  by  .01  in.  and  the  angle  A  by  1°.  Calculate  also 
the  differential  of  area  corresponding  to  the  same  increments. 

11.  The  distance  between  two  points  A  and  B  on  opposite  sides 
of  a  pond  is  determined  by  taking  a  third  point  C  and  measuring 
^C  =  90  ft.,  BC  =  110  ft.,  and  BCA  =  60°.  Find  approximately  the 
greatest  possible  error  in  the  computed  length  oi  AB  caused  by 
possible  errors  of  4  in.  in  the  measurement  of  both  AC  and  BC. 

12.  The  distance  of  an  inaccessible  object  A  from  a  point  B  is 
found  by  measuring  a  base  line  5C  =  100  ft., the  angle  CBA  =  a=  45°, 
and  the  angle  BCA  =  p  =  60°.  Find  the  greatest  possible  error  in 
the  computed  length  oi  AB  caused  by  errors  of  1'  in  measuring  both 
a  and  p. 

13.  The  equal  sides  of  an  isosceles  triangle  are  increasing  at  the 
uniform  rate  of  .01  in.  per  second,  and  the  vertical  angle  is  increas- 
ing at  the  uniform  rate  of  .01  radians  per  second.  How  fast  is  the 
area  of  the  triangle  increasing  when  the  equal  sides  are  each  2  ft. 
long  and  the  angle  at  the  vertex  is  45°  ? 

14.  Prove  that  the  rate  of  change  of  ^  =  \n{x  -\-^x^-\-  y^)  in  the 
direction  of  the  line  drawn  from  the  origin  of  coordinates  to  any 
point  P  (a;,  y)  is  equal  to  the  reciprocal  of  the  length  of  OP. 

15.  The  altitude  of  a  right  circular  cone  increases  at  the  uniform 
rate  of  .1  in.  per  second,  and  its  radius  increases  at  the  uniform  rate 
of  .01  in.  per  second.  How  fast  is  the  lateral  surface  of  the  cone 
increasing  when  its  altitude  is  2  ft.  and  its  radius  1  ft.? 


GENERAL  EXERCISES  193 

■^ ^  1 4-  a; 

16.  Given  z  =  tan"^— ^ h  taii"^ Find  the  general  expres- 
sion for  the  derivative  of  z  along  the  line  drawn  from  the  origin  of 
coordinates  to  any  point.  Find  also  the  value  of  this  derivative  at 
the  point  (1,  1). 

17.  In  what  direction  from  the  point  (3,  4)  is  the  rate  of  change 
of  the  function  z  =  kxi/  a  maximum,  and  what  is  the  value  of  that 
maximum  rate  ? 

18.  Find  a  general  expression  for  the  rate  of  change  of  the  func- 
tion u  =  e-v  sin  x  +  -  e-sv  sin  3  a;  at  the  point  ( ^ »  ^  )•  Find  also  the 
maximum  value  of  the  rate  of  change. 


CHAPTER  IX 

INTEGRATION 

65.  Introduction.  In  §§18  and  23  the  process  of  integration 
was  defined  as  the  determination  of  a  function  when  its  deriva- 
tive or  its  differential  is  known.    We  denoted  the  process  of 

integration  by  the  symbol  i;  that  is,  if 
f(x)dx^dF(x), 
then  Cf(x)  dx  =  F(x)  +  (7, 

where  C  is  the  constant  of  integration  (§18). 

The  expression  f(x)  dx  is  said  to  be  under  the  sign  of  inte- 
gration, and /(a:)  is  called  the  integrand.  The  expression  F(x}  -f  C 
is  called  the  indefinite  integral  to  distinguish  it  from  the  definite 
integral  defined  in  §  23. 

Since  integration  appears  as  the  converse  of  differentiation, 
it  is  evident  that  some  formulas  of  integration  may  be  found 
by  direct  reversal  of  the  corresponding  formulas  of  differentia- 
tion, possibly  with  some  modifications,  and  that  the  correctness 
of  any  formula  may  be  verified  by  differentiation. 

In  all  the  formulas  which  will  be  derived,  the  constant  C  will 

be  omitted,  since  it  is  independent  of  the  form  of  the  integrand; 

but  it  must  be  added  in  all  the  indefinite  integrals  found  by 

means  of  the  formulas.    However,  if  the  indefinite  integral  is 

found  in  the  course  of  the  evaluation  of  a  definite  integral,  the 

constant  C  may  be  omitted,  as  it  will  simply  cancel  out  if  it  has 

previously  been  written  in  (§23). 

The  two  formulas        •»  ^ 

/  cdu  =  c  I  du  (1) 

and      j{du  -\-dv-{-dw-{ )  =  jdu  +  fdv  -f  fdw  -|-  •  •  •      (2) 

194 


INTEGRAL  OF  u-  195 

are  df  fundamental  importance.     Stated  in  words  they  are  as 
follows : 

(!')  A  constant  factor  may  he  changed  from  one  side  of  the  sign 

of  integration  to  the  other. 

(2)    The  integral  of  the  sum  of  a  finite  number  of  functions  is 
the  sum  of  the  integrals  of  the  separate  functions. 

To  prove  (1),  we  note  that  since  cdu  =  d(^cu),  it  follows  that 


I  cdu  =  I  d  (cu)  =  cu  =  c  I  du. 


In  like  manner,  to  prove  (2),  since 

du-{-  dv  -{-dw  -\-  •  *  '  =  d(u-\-v  -\-w  -{-  '  *  '^y 
we  have 

I  (du  -{-  dv  -\- dw  -{-  -  '  ')  =  id(u-\-v  +  w-\-»''') 

=  Cdu  4-  Idv  4-  \dw  -\ . 

The  application  of  these  formulas  is  illustrated  in  the  follow- 
ing articles. 

66.  Integral  of  i/".    Since  for  all  values  of  m  except  m  =  0 

d(tr}  =  mu"'~^du, 
or  d(—j=u"'-^du, 

u"'~^du  =  — . 
m 

Placing  w  =  n  4-1,  we  have 

u-du  =  ^^  (1) 

for  all  values  of  n  except  n  =  —  1. 

In  the  case  n  =  —  l,  the  expression  under  the  sign  of  inte- 
gration in  (1)  becomes  — ?  which  is  recognized  as  c?(lnM). 

u 


Therefore 


/^  =  ln«.  (2) 


196  INTEGRATION 

In  applying  these  formulas  the  problem  is  to  choose  for  u 
some  function  of  x  which  will  bring  the  given  integral,  if  pos- 
sible, under  one  of  the  formulas.  The  form  of  the  integrand 
suggests  the  function  of  x  which  should  be  chosen  for  u. 

Ex.  1.    Find  the  value  of  j  (ax^  +  bx  -\ h  — )  dx. 

Applying  (2),  §  65,  and  then  (1),  §  65,  we  have 

I  I  ax^  +  bx  •] h  —\dx  =  j  ax^ dx  +  j  bxdx  -\-  j  -dx  ■{■  j  -^dx 

=  a  I  x'^dx  ■}■  b  I  xdx  -\-  c  j [■  k  j  x~^ dx. 

The  first,  the  second,  and  the  fourth  of  these  integrals  may  be  evaluated 

by  formula  (1)  and  the  third  by  formula  (2),  where  u  =  x,  the  results  being 

1         1  it 

respectively  -  ax%  -  hx% »  and  c  In  x. 

o  Jd  x 

Therefore    C lax''  ■\-  hx  ^-  ■{■  !^dx  =  \ax^  ■\-\hx'^  ^-  c\nx  --  -V  C. 
•/   \  a:      xV  3  2  x 

Ex.  2.  Find  the  value  ofj(x^+  2)xdx. 

If  the  factors  of  the  integrand  are  multiplied  together,  we  have 

f(x^+2)xdx=j(x^+2x)dxy 

which  may  be  evaluated  by  the  same  method  as  that  used  in  Ex.  1,  the 
result  being  ^  x*  -\-  x^  •}■  C. 

Or  we  may  let  x^-^2  =  u,  whence  2xdx  =  du,  so  that  xdx—\  du.   Hence 


f(x^+  2)xdx=f^udu  =  ifudu 


2    2 
=  i(a:2+2)2+C. 

Comparing  the  two  values  of  the  integral  found  by  the  two  methods  of 
integration,  we  see  that  they  differ  only  by  the  constant  unity,  which  may 
be  made  a  part  of  the  constant  of  integration. 

Ex.  3.  Find  the  value  of   f(ax^  +  2  hxy(ax  +  h)dx. 

Let  ax^  +  2hx  =  u.  Then  (2  ax  +  2  b)  dx  =  du,  so  that  (ax  +  b)  dx  =  |  du. 

Hence       f  (ax^  +  2bxy (ax  -{■  b)dx  =   f  ^u^du 

=  l(ax^  +  2bxy+C. 


INTEGRAL  OF  ir  197 

Ex.4.  Find  the  value  of   rli^f±^. 

As  in  Ex.  3,  let  ax^  -\-2bx  =  u.  Then  (2ax  +  2b)dx  =  du,  so  that 
(ax  +  b)dx  =  Idu. 

'  Hence  ri(ax  +  b)dx^r2_du^^r±i 

J     ax^  +  2bx        J      u  J    u 

=  2  In  M  +  C 

=  2  In  (ax2  +  2  6a:)  +  C 

=  ln(aar2  +  2  6x)2+C. 

Ex.  5.  Find  the  value  of   f  (e«^  +  b^e^dx. 

Lete«*  +  6  =  M.    Then  c«^a</a:  =  ^/u. 

Hence  ("(e-*  +  6)2e«^t/x  =    fti*  — 

a  •/ 
d  a 

o  a, 

If  the  integrand  is  a  trigonometric  expression  it  is  often  pos- 
sible to  carry  out  the  integration  by  either  formula  (1)  or  (2), 
This  may  happen  when  the  integrand  can  be  expressed  in  terms 
of  one  of  the  elementary  trigonometric  functions,  the  whole 
expression  being  multiplied  by  the  differential  of  that  function. 
For  instance,  the  expression  to  be  integrated  may  consist  of  a 
function  of  sin  2:  multiplied  by  co^xdx^  or  a  function  of  cos  a; 
multiplied  by  {—^mxdx),  etc. 

Ex.  6.  Find  the  value  of  i  Vsinx  cos^xdx. 

Since  </(sina:)  =  cosa:c?x,  we  will  separate  out  the  factor  cosxdx  and 
express  the  rest  of  the  integrand  in  terms  of  sin  x. 

Thus  Vsina:  cos^a:c?a:  =  Vsin x (1  —  sin^a:)  (cosxdx). 

Kow  place  sin  a:  =  m,  and  we  have 

r  Vsin  a;  cos'a;£?a;  =  \vi  (1  —  m*)  du 
—  j  (ui—  u^)du 

=  ^j  sinta:(7-  3  sin^a:)  +  C 


198  INTEGRATION 

Ex.  7.    Find  the  value  of  j  sec^2xdx. 

Since  c?(tan  2  x)  =  2  860^2  xdx,  we  separate  out  the  factor  860^2  xdx  and 
express  the  rest  of  the  integrand  in  terms  of  tan  2  x. 

Thus  sec^  2xdx  =  sec*  2  x  (sec^  2  x  dx) 

=  (1  +  tan2  2  xy(sec^2xdx) 

=  (1  +  2  tan2  2  x  +  tan*  2  x)  (sec22  a;c?a:). 

Now  place  tan  2x  =  u,  and  we  have 

fsec^2xdx  =  ^f  (1  +  2  u^+ u*)du 

=  ^tan 2  a:  +  ^tan32  x  +  j^jjtdLJi^2  x  ■{■  C. 

EXERCISES 

Find  the  values  of  the  following  integrals : 

J   \  xl  J    6^«*  +  tan  ax 

3.   r/^v^ — \\a,.         13.  r   ^"^^^    dx. 

J  \  x-Sxj  J  l  +  cosaa; 

4.    C^^\^^\x.  14.    Ceos^2xsm2xdx. 

— — -•  15.    I  sin^SiccosSa-t/o?. 

6.  i  {x^  +  lfxdx.  16.    I  sin(x  +  2)cos(a;  +  2)c?a;. 

7.  I  Voj^H-  Ax^dx.  17.    I  cos^3x  sin  3xc?a;. 

8- J^^TTfTe-  18.  Jsec^S^c^x. 

9-    r^  +  ^Q^^^  dx.  19.    rctnV2a;+l)cscV2a^+l)^a!. 

/\ (»os  X  r 
; -idx.                   20.    I  sin«(2x-3)c?a;. 


ALGEBRAIC  INTEGRANDS  199 

67.  Other  algebraic  integrands.  From  the  formulas  for  the 
differentiation  of  sin"^t«,  tan~^w,  and  sec~'M,  we  derive,  by  re- 
versal, the  corresponding  formulas  of  integration: 

du 


J 


Vl  -  u' 
du 


=  sin  ^w, 


/ 


=  tan~^w, 

/du 


du 


These  formulas  are  much  more  serviceable,  however,  if  u  is 


u 


replaced  by  -  (a  >  0).    Making  this  substitution  and  evident 

reductions,  we  have  as  our  required  formulas 

du  .     ,u 


f- 


V«2-  u^  a 


(1) 


du         1 ,      _^u  ,^. 

-  tan  ^  - »  (2) 


and 


/du  1       _^u^  „ 


u^  +  a^      a  a 

du  1       _j  w 


u 


Referring  to  1,  §  47,  we  see  that  sin~^-  must  be  taken  in  the 

first  or  the  fourth  quadrant;    if,  however,  it  is  necessary  to 

u 
have  sin~^  -  in  the  second  or  the  third  quadrant,  the  minus  sign 

u 
must  be  prefixed.   In  like  manner,  in  (3),  sec~^  -  must  be  taken  in 

the  first  or  the  third  quadrant  or  else  its  sign  must  be  changed. 


/dx 
—  ■ .  Letting  2  a;  =  w,  we  have  du  =  2dx', 

V9-4a:2 

/dx        _  r    ^du 


whence  dx  =  ^  du,  and  ^  ^     '*  ^ 


V9  -  4  a;2     -^  V9  -  m^ 


_  1    /»      du 
~  2  J  V9  -  w2 


1    .      ,2a: 


=  -  sin 


2  3 


^  +  C. 


200  INTEGRATION 

dx 


Ex.  2.  Find  the   value  of   f — ,  If   we    let  Vs  x  -  u,   then 

^    ^  xVSx^-i 
— zzdu,  and 

V3 

dx  r        du 


_  ^    -   xvo 

<fu  =  V3  rfx :  whence  dx  =  — -=  rfw,  and 

V3 


a:  V3  a:2  —  4      "^  w  Vm'^  _  4 
=  |sec-i|+C 


r/a: 


Ex.  3.  Find  the  value  of  J  '  ,- 

*'  V4a:  —  x^ 

Since  V4  x  —  x"^  =  V4  —  (.r  —  2)^,  we  may  let  m  =  a:  —  2 ;  whence  dx  —  du,  and 


dx  r      du 


/dx  r 

a/4.  7-  _  7-2      -^ 


V4  X  —  a;'-^      "^  V4  —  w2 
=  sin-i  ^  +  C 

Ex.  4.  Find  the  value  of  J^-^-^-j-^ . 

We  may  first  write  the  integrand  in  the  form 

1  1  1  1 

2'x2+ix+t       2*(a  +  i)=^+n' 

and  let  M  =  X  +  i-    Then  du  =  dx, 

r  dx  _  1  r  dx 

^"""^  J  2x^+3x  +  5~2J  (x+^y+U 


_  1   r     du 

"2J  u2+f^, 


=  -.  tan-'-_=+C 

V3I  V3I 

4  4 

=  -7=rtan-*— ^+  C 
V3I  V31 

=  — ==-tan-^ — —=r-  +  Cr 
V3I  V3I 


ALGEBRAIC  INTEGRANDS  201 


Ex.  5.    Find  the  value  of  |  — dx. 

Separating  the  integrand  into  two  fractions 
5ar  2 


2  a;2  +  3      2  x^  +  3 
and  using  (2),  §  65,  we  have 

5  X  —  2   ,         r  5xdx         r    2dx 


/o  X  —  2       _  r  bxdx   _  r    "JLdx 
2x2+3    ^~J  2x2+3      J  2x2  + 


If  we  let  M  =  2  x2  +  3,  then  du  =  4  xdx 

,  r  5  xdx        5  rdu      5,  5 ,     .^   „  ,  „. 

and  /  — — =  -  /  —  =  -  In  M  =  -  In  (2  x2  +  3)  ; 

^2x2+3      ^J    u       4  4      '^  ^' 

and  if  we  let  u  =  V 2  x,  then  du  =  V2  Jx 

and       r-^=V2r^  =  V2.^_tan-A=:^tan-^ 
J  2x2+ 3  Jw2+3  V3  V3        3  3 

Therefore  f^T^  dx  =  5 In (2x2+  3)  _  2L?tan-^^^. 

^2x2+3  4^  ^         3  3 


Ex.  6.  Find  the  value 


°^x: 


Vs    c?x 


/_ 


X2+1 

-^—^  =rtan-ixV*  =tan-iV3-tan-i(-l). 

1      X2+1  l-  -J-1  ^  ^  . 

There  is  here  a  certain  ambiguity,  since  tan-^V3  and  tan-i(— 1)  have 
each  an  infinite  number  of  values.  If,  however,  we  remember  that  the  graph 
of  tan~^x  is  composed  of  an  infinite  number  of  distinct  parts,  or  branches 
(Fig.  56,  §  46),  the  ambiguity  is  removed  by  taking  the  values  of  tan-^  V3 
and  tan-^(— 1)  from  the  same  branch  of  the  graph.    For  if  we  consider 

dx 

tan~^&  —  tan-^a  and  select  any  value  of  tan~^a,  then  if  Z>  =  a, 


X 


X2+1 

tan~^6  must  be  taken  equal  to  tan~^a,  since  the  value  of  the  integral  is 
then  zero.  As  b  varies  from  equality  with  a  to  its  final  value,  tan-^6  will 
vary  from  tan-^a  to  the  nearest  value  of  tan- ^6. 

The  simplest  way  to  choose  the  proper  values  of  tan-^Z;  and  tan-^a  is 

to  take  them  both  between  —  -  and  -  •    Then  we  have 


rVs    dx     _  ^  _  /     ir\  _  7j 
J-i  x2  +  l~3      V     4/       li 


12  ' 


The  same  ambiguity  occurs  in  the  determination  of  a  definite  integral 
by  (1),  but  the  simplest  way  to  obviate  it  is  to  take  both  values  of  sin-^  — 


between and  - .   The  proof  is  left  to  the  student. 

2  2  ^ 


202  INTEGEATION 

EXERCISES 

Find  the  values  of  the  following  integrals  : 

dx  r         dx 


1. 


/dx  r 


dx  r  dx 


/dx  r 

/dx  '             r d^ 

r     d^ r3^  +  ii 

dx r2x-\-^ 

V2  +  4a;-3a;2*  ^^*  J  V4  -  x' 

/a 

dx  r^       dx 

^     r    dx  r^  dx 


.  4 

3        6^0; 


10, 


9 
dx  r^  dx 


J   V6a;-4x2  r 


^a;V4ic2_9 


68.  Closely  resembling  formulas  (1)  and  (2)  of  the  last  section 
in  the  form  of  the  integrand  are  the  following  formulas : 

r-^  =  ln(«+Vi?T7^),  (1) 

r-^  =  ln(«+Vi?3^),  (2) 

/du     _   1    ,    u—  a  g 


and 


These  formulas  can  be  easily  verified  by  differentiation,  and 
this  verification  should  be  made  by  the  student. 


ALGEBKAIC  INTEGRANDS  203 


/dx 

Letting  '\/2  x  =  u,  we  have  du  =  V2  dx ;  whence  dx  =  — =•  </m,  and 

V2 


/ 


V2  x2  -  3      J    Vm^-3 
_    1      /*      rft^ 

a/9  »^    a/,/2  _ 


V2  *^  Vw- 

i  In  [m  +  Vw-^  -  3]  +  C 

V2 


-^ln[zV2  +  V22-2-  3]  +  C. 

V2 


Ex.  2.  Find  the  value  of  ' 


.  r         dx 
^  V3  x^  +  4 


As  in  Ex.  4,  §  67,  we  may  write  the  integrand  in  the  form 
111  1 


V3    Va:2  +  4  X      V3    y/(x  +  f  )2  -  f 
and  let  m  =  a:  +  § ;  whence  du  =  dx. 


/dx       _  1    r         dx 
V3  x2  +  4  X  ~  V3  -^  V(x+  i)2- 


f7u 


VS*^    Vm2  -  I 


V3 

V3 

=  iln(3  X  +  2  +  V9x2  +  12ar)  H-iT, 
V3 


where  C  =  i  In  3  +  iT. 
V3 


2  x2  +  X  - 
Writing  the  integrand  in  the  form 
1  1  1 


2    x^+^x-V-      2    (x+i)2--i^ 
we  let  M  =  X  +  I ;  whence  du  =  t/x. 


204  INTEGRATION 

_  1   r      du 
-  2  J  m2  _  j^ 


2    2(-V)      w  +  V 
11      x  +  3 


11        X  +  o 


where  C  =  ^^  ^^  2  +  iT. 


EXERCISES 

Find  the  values  of  the  following  integrals  : 

/dx                        , ,    r    ^^ 
.  11.      I    r— ; z 

^*  J   V9x^-1*  •  J  a;^- 3x4-1" 


+  5a;-2 


3.  r^^==-  13.  r-, 

•     J   V3a;^-4  J  ^ 

/<^a;  r           dx 

■\/x^  +  2x  '  J  4.x^-2x- 

r        dx  r^    dx 

^'  J  ■\^3x'-j-2x-^3  '  Ji   Vx^  -  4* 

/dx  r^        dx 

j  2x^-1*  ^^- j^    V9^ 


6x  —  3 


/dx  pi  dx 

S^^^^'  ^^'  X    V2x«-2x  +  l 

/dx  r^         dx 

x^4-4x*  ^^*  J^   2x2- X- 3' 


TRIGONOMETRIC  FUNCTIONS  205 

69.  Integrals  of  trigonometric  functions.  Of  the  following  for- 
mulas for  the  integration  of  the  trigonometric  functions,  each 
of  the  first  six  is  the  direct  converse  of  the  corresponding  for- 
mula of  differentiation  (§  44),  and  the  last  four  can  readily  be 
verified  by  differentiation,  which  is  left  to  the  student. 

/  sin  udu  =  —  cos  u,  (1) 

I  cos  udu  =  sin  u,  (2) 

I  see^uda  =  tan  u,  (3) 

I  csc^ udu  =  —  ctn  w,  (4) 

I  sec  u  tan  udu  =  sec  w,  (5) 

j  CSC  u  ctn  udu  =  —  esc  w,  (6) 

/  tan  udu  =  In  sec  u,  (7) 

I  ctn  udu  =  In  sin  w,  (8) 

I  sec  udu  =  \n (sec  u  +  tan w),  (9) 

/  CSC  w(?w  =  In  (esc  u  —  ctn  it).  (10) 

Ex.  1.    Find  the  value  of  Jsin  7xdx. 

If  we  let  u  =7 X, 

then  du  =  7  dx'j 

whence  dx  =  \  du, 

and  C sin7 X dx  =  j  sin  u(]^du) 

=  if-  I  sinwrfu 

=  —  \  cos  M  +  C 

=  — }cos7a;  +  C. 


206  INTEGRATION 

Ex.  2.    Find  the  value  of  Tsec  (2  x  +  1)  tan  (2  a;  +  1)  dx. 

If  we  let  u  =  2  X  -{■  1,  then  du  =  2  dx, 

and   jsec(2x  +1) tan (2  a:  ■\-l)dx  =  ^  C sec  u  tan  udu 

=  ^  sec  M  +  C 

=  1  sec  (2  X  +  1)  +  C. 

Often  a  trigonometric  transformation  of  the  integrand  facili- 
tates the  carrying  out  of  the  integration,  as  shown  in  the 
following  examples: 

Ex.  3.    Find  the  value  of  j  cos^axdx. 

Since  cos^aa;  =  I  (1  +  cos  2  ax), 

jcos^axdx  =  C(^  +  ^cos  2  ax)dx 

—  ^  Cdx  +  \  fcos  2  cixdx 

=  -  X  +  - —  sin  2  ax  +  C, 
2  4a 

the  second  integral  being  evaluated  by  formula  (2)  with  u  =  2  ax. 

Ex.  4.    Find  the  value  of  T  Vl  +  cos  xdx. 

Since  cosx=2cos'^ 1, 

2 

Vl  +  COS  a;  =  "V  2  cos  - , 

and  f^^  "^  cos  a:  (/a;  =  T  V2  cos  -  dx 

=  V2  Ccos  ^  dx 

=  2V2sin|  +  C. 
Ex.  5.    Find  the  value  of  ftan^S  xdx. 
Since  tan^Sa:  =  sec^S  a:  —  1, 

Ttan^S  xdx=  /"(sec^S  x-l)dx 

=  Csec^d  xdx—  Cdx 

=  ^  tan  3  ar  —  x, 
the  first  integral  being  evaluated  by  formula  (3)  with  w  =  3  a:. 


TRIGONOMETRIC  FUNCTIONS  207 

EXERCISES 

Find  the  values  of  the  following  integrals  : 

1.  jsm(Sx-2)dx.  13.    Ceos^^dx. 

2.  I  cos  (A  —  2  x)  dx,  14.    j  Isin  - -^  cos -j  dx. 

3.  rsec(3ic-l)tan(3ic-l)6^x.  15.    j  (sec  ^  -  t^n -~J  dx. 

4.  Csec^^dx.  16.  J  sin^jcos^jc^x. 
jt^J-§dx.  17.f^l+cos^dx, 

I  ctn  5xdx.  '  J 


5. 


18.    I   Vl  —  cos  4:xdx. 
sin  Sxdx. 


7.  Ccsc(2x-^S)dx.  ^^-  J^    ^^ 

^      r       ic    ,    ic  -  20.    /   ^tan-c?ic. 

8.  I  csc-ctn-c?a:.  J^  2 

r  21.      f^h3iTl'(x-\-^)dx. 

9.  I  sec(4ic  +  2)^ic.  Jo  ^         ^z 

r  22.  r 

LO.    /  080^^(3-  2x)dx.  Jj 


sec  2  0?  dx. 


n.   p-2i2x^^  23.   rw(|  +  f)<;.. 

J     since  .  *^-4 

12.    I  siii^^dx.  24.    |    Sec^2a:c?a:. 

70.  Integrals  of  exponential  functions.    The  formulas 

and  I  a''du  =  - — a"  (2) 

J  In  a 

are  derived  immediately  from  the   corresponding  formulas  of 
differentiation. 


'8 

INTEGRATION 

E: 

K.  1.  Find  the  value  of  /  e^^dx. 

If 

we  let  3  :r  =  u,  we  have 

Je^^'dx  = 

:  ^fe'^du 

:^e»  +  C 

E: 

K.  2.  Find  the  value  of  f^dx. 
•^    x^ 

:i63-4-C. 

If 

X  -            -                        1 

we  place  V  5  =  5^  and  let  -  =  u, 

we  have 

J  I.  ^^- 

..^fb^du 

Ino 

Ino 

EXERCISES 

Find  the  values  of  the  following  integrals : 

1.    Ce^^^^dx.  5.    [{e'^+e-'^fdx.  9.    j    lO^dx. 

e^xdx.  6.    / dx.  10.    /     2^'''^^mxdx. 

3.  C{e  +  x'^)dx.  7.    f^rz\^^'  ^^'    r  ^'""'^^• 

4.  fe^  +  ^^c^  +  ^^tZx.       8.    f  \e^  +  e'Vdx.        12.     |    -f^-^dx. 
J  Jo  Jo    e"+«-" 

71.  Substitutions.  In  all  the  integrations  that  have  been 
made  in  the  previous  sections  we  have  substituted  a  new  vari- 
able u  for  some  function  of  x,  thereby  making  the  given  integral 
identical  with  one  of  the  formulas.  There  are  other  cases  in 
which  the  choice  of  the  new  variable  u  is  not  so  evident,  but 
in  which,  nevertheless,  it  is  possible  to  reduce  the  given  integral 
to  one  of  the  known  integrals  by  an  appropriate  choice  and  sub- 
stitution of  a  new  variable.  We  shall  suggest  in  this  section  a  few 
of  the  more  common  substitutions  which  it  is  desirable  to  try. 

I.  Integrand  involving  powers  of  a-\-hx.  The  substitution  of 
some  power  of  z  for  a-\-hx  is  usually  desirable. 


SUBSTITUTIONS  209 

Ex.  1.  Find  the  value  of  f     ^'"^^      . 

Here  we  let  1  +  2  x  =  2^ ;  then  a:  =  ^  (^^  -  1)  and  dx  =  ^  z^dz. 

Therefore  r_J^dx      ^  3   r ^^,  -2z'-i-z)dz 

•^  (l+2a:)i      8^ 

=  3i(T2^(5  ^6  - 16  ^3  +  20)  +  a 

Replacing  z  by  its  value  (1  +  2  a:)^  and  simplifying,  we  have 

j_^^^    3  I  2a:  +  20x2)  +  C. 

••^(l+2ar)^      320 

II.  Integrand  involving  powers  of  a-\-  hx*".  The  substitution 
of  some  power  of  z  for  a  +  bx""  is  desirable  if  the  expression 
under  the  integral  sign  contains  af'^dx  as  a  factor,  since 
c?  (a  +  hx"")  =  hnaf  ~  ^  dx. 


Ex.  2.  Find  the  value  of  f^l^^-L^dx. 
We  may  write  the  integral  in  the  form 

(xdx) 


rVx^  +  g^ 


and  place  x^  -^r  a^  =  z^.   Then  xdx  =  zdz,  and  the  integral  becomes 

J  z^-a"     J   \        z^-aV  2      s  +  a 

Replacing  z  by  its  value  in  terms  of  x,  we  have 

I  rfx  =  Va;2  +  a-'  +  -  In    ,  +  C. 

^  2       v^M^  +  a 

Ex.  3.  Find  the  value  of  fx^(l  +  2x^)^dx. 
We  may  write  the  integral  in  the  form 

fx\l  +  2x^)i(x^dx), 
and  place  1  +  2x^  =  z^.   Then  x^dx  =  ^zdz,  and  the  new  integral  in  z  is 
^f(z^  -  z^)dz  =  ^z\^z^  -  5)  +  C. 
j^      Replacing  z  by  its  value,  we  have 


210  INTEGKATION 


III.   Integrand    involving   ^a^—x\     If    a    right    triangle    is 
constructed  with  one  leg  equal  to  x  and 
with  the  hypotenuse  equal  to  a  (Fig.  83), 
the  substitution  x  =  a  sin  <^  is  suggested. 


Ex.  4.    Find  the  value  of  f  Va^  -  x'^dx. 

Let  a;  =  a  sin  <^.    Then  dx  =  a  cos  <f)  dcf>  and,  from  the  triangle,  Va-  —  x'^ 
—  a  cos  <^. 

Therefore  fVa^—  x^dx  =  a^  Ccos^cf)dcf> 

=  ^a^  C  (1  + cos  2  (fi)d<f> 

=  la^(<}>  +  I  sin  2  <^)  +  C, 

x 
But  <^  =  sin-  ^  -  , 

a 

and  sin  2  ^  =  2  sin  <^  cos  <^ 


2  x  V  a2  _  a;2 


a' 


X  V  flt^  —  37^ 

for,  from  the  triangle,  sin  cf>  =  -  and  cos  ch  = 

a  a 

Finally,  by  substitution,  we  have 


r  Va2  _  xhlx  =  2  (-^  ^^^  "  ^'^  +  a^sin-i-j  +  C. 


IV.   Integrand    involving    y/x'^  +  a^    If    a    right    triangle    is 
constructed  with  the  two  legs  equal  to  a: 
and  a  respectively  (Fig.  84),  the  substitu-  v/o^>" 

tion  x  =  a  tan  <^  is  suggested. 


Ex.  5.  Find  the  value  of  / 


dx 


(x^  +  a2)i 


Let  X  =  a  tan  <^.    Then  dx  =  a  sec^ff>d<f>  and,  from  the  triangle,  Vx^  +  a?- 
—  a  sec  <f>. 

Therefore  / =—  f — ^  =  —  f  cos  (f>dd>  = —  sin  th  +  C. 

(x2  +  a^)l      «  *^  sec  <f)      a^  J  a^ 

But,  from  the  triangle,  sin  (f>  —  — — = ;  so  that,  by  substitution, 

Vx^  +  a2 


SUBSTITUTIONS  211 

•  ^^ 

V.   Integrand    involving    y/x^—  (f.    If    a    right    triangle    is 
constructed  with  the  hypotenuse  equal  to 
X  and  with  one  leg  equal  to  a  (Fig.  85),  J^ 

the  substitution  ic  =  «  sec  <^  is  suggested. 

Ex.  6.  Find  the  value  oifx^Vx^  -  d^dx. 

Let  X  =  a  sec  <^.    Then   dx  =  a  sec  <^  tan  <f>  d<f>  and,  from   the  triangle, 

Vx^  —  a^  =  a  tan  <f>. 

Therefore  |  x^  Vx'-^  —  d^dx  =  a^  j  tam^ffy  sec*  (f>d<f> 

=  a^  j  (tan^*^  +  tsin*<f))sec^<f)d<f> 


Va:2  —  a^ 

But,  from  the  triangle,  tan  <f)= ;  so  that,  by  substitution,  we  have 

a 

fx^Vx^-a^dx  =  T^(2a^  +  S  x^)  V(x^  -  d^f  +  C. 

We  might  have  written  this  integral  in  the  form  Cx^-y/x^  —  a^(xdx^  and 
solved  by  letting  z^  =  x^  —  d^. 

72.  If  the  value  of  the  indefinite  integral  is  found  by  substitu- 
tion, the  evaluation  of  the  definite  integral  I  f(x)dx  may  be 

performed  in  two  ways,  differing  in  the  manner  in  which  the 
limits  are  substituted.  These  two  ways  are  shown  in  the  solutions 
of  the  following  example : 

Ex.    Find  f'Va^-x^dx. 
Jo 

By  Ex.  4,  §71, 

JVa^-x^dx  =  Ux  Va2  -  ^2  +  a^sin-i-)  +  C. 

Therefore      f^Va^  -  x^dx  =  l^(v  Va^  -  x^  +  a^sin-^-XV 

=  i/a  V^2T^  4-  a^sin-i-) 

-  i/o  Va2-0  +  a^sin-i-) 


tra' 
4 


212  mTEGEATIO:^^ 

Or  we  may  proceed  as  follows  :  Let  x  =  a  sin  <^.    When  x  =  0,  <^  =  0 ; 

and  when  a:  =  a,  ^  =  -,  so  that  <f>  varies  from  0  to  -  as  a;  varies  from  0  to  a. 
Accordingly,  "  ^ 


=  g(*+|sin2.^)] 


2 
-  _  0 

_  Tra^  / 

The  second  method  is  evidently  the  better  method,  as  it  obviates  the 
necessity  of  replacing  z  in  the  indefinite  integral  by  its  value  in  terms  of 
X  before  the  limits  of  integration  can  be  substituted. 


EXERCISES 

Find  the  values  of  the  following  integrals : 

r  x^dx  r  x'dx  r^     dx 


(4 

dx  \\     r^      dx 


J   Wx-l  J  (3-x2)^  J-i 


3.    fx(2x-S)^dx.      8.    f ^=.         13.    r 

J  J  xWl-\-4.x'  Ji 


i^+iy 


5.    /  — 7-  10.    /    x-\/2x-3dx.    15.    r  xVT^^hlx. 

73.  Integration  by  parts.  Another  method  of  importance  in 
the  reduction  of  a  given  integral  to  a  known  type  is  that  of 
integration  hy  parts,  the  formula  for  which  is  derived  from  the 
formula  for  the  differential  of  a  product, 

d(uv)  =  udv  +  vdu. 
From  this  formula  we  derive 

uv=  I  udv  +  I  vduy 
which  is  usually  written  in  the  form 


j  udv  =  uv  —  I  V du. 


INTEGRATION  BY  PAETS  213 

In  the  use  of  this  formula  the  aim  is  evidently  to  make  the 
original  integration  depend  upon  the  evaluation  of  a  simpler 
integral. 

Ex.  1.  Find  the  value  of  Cxe'^dx. 

If  we  let  a:  =  M  and  e'dx  =  dv^  we  have  du  =  dx  and  r  =  c*. 
Substituting  in  our  formula,  we  have 

ixe'^dx  —  xeF—  \  e^dx 
=  xe^-  e^  +  C 

=  (X  -1)€^+  p. 

It  is  evident  that  in  selecting  the  expression  for  dv  it  is  desirable,  if 
possible,  to  choose  an  expression  that  is  easily  integrated. 


Ex.  2.  Find  the  value  of  isin.-'^xdx. 


dx 
Here  we  may  let  sin-^  a:  =  u  and  dx  =  dv,  whence  du  =  and  v  =  x. 

Substituting  in  our  formula,  we  have  V 1  —  x 


I  sin~^xdx  =  X  sin-^x  —  i  - 


xdx 


VI -x2 
=  X  sin-ix  +  Vl  -  a;2  +  C, 
the  last  integral  being  evaluated  by  (1),  §  66. 

Sometimes  an  integral  may  be  evaluated  by  successive  inte- 
gration by  parts. 

Ex.  3.  Find  the  value  of  fx^e^dx. 

Here  we  let  x^  =  u  and  e'^dx  =  dv.   Then  du  =  2  xdx  and  v  =  e^. 

Therefore  Cx^e^dx  —  a;V  _  2  Cxef^dx. 

The  integral  I  xe^dx  may  be  evaluated  by  integration  by  parts  (see  Ex.  1), 
so  that  finally 

fx^e^'dx  =  a:V  -  2  (a:  -  1) e^  +C  =  ^^(x^  -  2  ar  +  2)  +  C. 

Ex.  4.  Find  the  value  of  j  e*^  sin  hxdx. 
Letting  sin&x  =  u  and  e"^(/x  =  dv,  we  have 

/e"^  sin  hxdx  =  -  e^  sin  bx I  e°^  cos  bxdx. 
a                      a  J 


214  INTEGEATION 

In  the  integral  fe"^  cos  hxdx  we  let  cos  bx  =  u  and  e"^dx  —  dv,  and  have 

/e«-»  cos  hxdx  =  -  e^  cos  bx  +  -  (  e"^  sin  bx dx. 
a  a  J 

Substituting  this  value  above,  we  have 

/e^  sin  bxdx  —  - e'^  sin  6x ( -  e'^  cos  6x  +  -   |  e«^  sin  &a:</;r ). 
a                       a\a  a  J  ) 

Now  bringing  to  the  left-hand  member  of  the  equation  all  the  terms 
containing  the  integral,  we  have 

(J2\    ^                           1                        h 
1  +  -i;  I   I  ^""^  sin  bxdx  =  -  e"^  sin  bx e«^  cos  Jar, 


whence  /  e"^  sin  bxdx  = 

J  a^  + 


g"^  (g  sin  bx  —  b  cos  ftx) 
Ex.  5.  Find  the  value  of  /  Vx^  +  a^dx. 


Placing  Vx^  +  a^  =  m  and  i/x  =  dv,  whence  du  =  — ==  and  v  =  a:, 
we  have  ^-^  "^  ^ 

rVx^  +  d^dx  =  X Vx'  +a'-  f     ^!l^ .  (l) 

Since  a;^  =  (a:^  +  a^)  —  a^,  the  second  integral  of  (1)  may  be  written  as 

/(x^  +  g^)  dx  _    2  f      dx 
V^M^^  »^  Va;2  -f  a2* 

which  equals  /  Va:^  -\-  a^dx  —  a^  i     , 

•^  -^  Vx2  +  g2 

Evaluating  this  last  integral  and  substituting  in  (1),  we  have 
fVx'  +  g^t/a;  =  x  Va;^  +  g2  -fVx^  +  gVa;  +  g2  In  (a:  +  Va;^  +  g2), 
whence     JVar^  +  a^dx  =  |  [a:  Vx2  +  a^  +  a^  In  (a:  +  Va;^  +  g2)]. 

74.  If  the  value  of  the  indefinite  integral  /  f(x')  dx  is  found 
by    integration    by   parts,    the    value    of   the    definite    integral 

Jf   f(x)dx  may  be  found  by  substituting  the  limits  a  and  5,  in 
a 

the  usual  manner,  in  the  indefinite  integral. 


mTEGRATION  BY  PARTS  215 

Ex.  Find  the  value  of  f^  x^sinxdx. 
Jo 

To  find  the  value  of  the  indefinite  integral,  let  x^  =  u  and  sin  zdx  =  dv. 
Then  /  x^ sin xdx  =—  x^ cos x  +  2  j  x  cos x dx. 

In  I  X  cos  xdx,  let  X  =  M  and  cos  xdx  =  dv. 

Then  I  x  cosxdx  =  x  sinx  —  j  sinxdx 

=  a:  sin  ar  +  cos  x. 
Finally,  we  have 

j  x^sinxdx  =—  ar^cosa;  +  2a:  sin  a:  +  2cosx  +  C. 

Hence  C^x^  sin  a:  Jx  =    —  x^  cos  x  ■{■  2x  sin  x  +  2  cos  x  I 

=  TT  -  2. 

The  better  method,  however,  is  as  follows:  ^ 

lif(x)dx  is  denoted  by  udv,  the  definite  integral   /  f(pc)dx 

udv,  where  it  is  understood  that  a  and  b 

are  the  values  of  the  independent  variable.    Then 

/    udv  =  [uv^  —  /    vdu. 

\J  a  J  a 

To  prove  this,  note  that  it  follows  at  once  from  the  equation 
^uv\  =  j    dQa,v)—  I    (udv  ■\-vdu)=  I    udv  +   I    vdu. 

xJ  a  tj  a  %J  a  *J  a 

Applying  this  method  to  the  problem  just  solved,  we  have 
r  2  a;2  sin  a;cfx  =  j  —  x^  cos  x\  ■{■  2  C'^  x  coaxdx 

IT 

=  2  C^ X  cos X dx 

=    2  a:  sin  a:    —  2  C^  ainxdx 

n 

=  IT  +    2  cos  x 
=  7r-2. 


1  X 

e'^dx. 


2xdx. 


216  INTEGRATION 

EXERCISES 

Find  the  values  of  the  following  integrals : 

1.  /  xe^^'dx.  5.  I  ccsec~^2icc?x.  9.  I  x 

2.  I  x^e^^'dx.  6.  I  (In  sinic)cosicc?ic.  10.  /  x'^lnxdx 

3.  I  GOS~'^xdx.  7.  I  e^'^GOSxdx.  11.  J  sin~^ 

4.  I  tan~^3a!cZa!.  8.  j  xeos'^-dx.  12.  J  *a;  cos  2arc?a;. 

75.  Integration  of  rational  fractions.  A  rational  fraction  is  a 
fraction  whose  numerator  and  denominator  are  polynomials.  It 
can  often  be  integrated  by  expressing  it  as  the  sum  of  partial 
fractions  whose  denominators  are  factors  of  the  denominator 
of  the  original  fraction.  We  shall  illustrate  only  the  case  in 
which  the  degree  of  the  numerator  is  less  than  the  degree  of  the 
denominator  and  in  which  the  factors  of  the  denominator  are 
all  of  the  first  degree  and  all  different. 

Ex.  Find  the  value  of  r_£l±il£+ll  ^^, 
J  {x  +  3)  (x'  -  4) 

The  factors  of  the  denominator  are  a:  +  3,  a:  —  2,  and  x  +  2.  We  assume 

(x  +  3)  (a;2  -  4)      x  +  3      x-2      x  +  2*  ^  ^ 

where  A ,  B,  and  C  are  constants  to  be  determined. 

Clearing  (1)  of  fractions  by  multiplying  by  (x  +  3)  (x^  —  4),  we  have 

x^+llx  +  U  =  A(x-2)(x  +  2)+B(x-{-S)(x  +  2)-\-C(x  +  d)(x-2),  (2) 

or    x''+llx  +  U=(A+B  +  C)x^+  (5B  +  C)x  +  {-4:A-\-QB-6C).  (3) 

Since  A,  B,  and  C  are  to  be  determined  so  that  the  right-hand  member 
of  (3)  shall  be  identical  with  the  left-hand  member,  the  coefficients  of  like 
powers  of  x  on  the  two  sides  of  the  equation  must  be  equal. 

Therefore,  equating  the  coefficients  of  like  powers  of  x  in  (3),  we  obtain 
the  equations  A  +  B+C  =  l, 

5B+C  =  11, 
-4^  +  65-6C  =  14, 
whence  we  find  A  =  - 2,  B  =  2,  C  =  1. 


RATIONAL  FRACTIONS  217 


Substituting  these  values  in  (1),  we  have 
x^  +  Ux  +  U  2       .      2 


(x  +  3)  (x2  -  4)  x  +  3      x-2      x  +  2 

,     r  x^  +  llx+U     .  ridx    ^     rldx    ^     n    dx 

""M(.  +  3)(.^-4)^"=-J^T3  +  /j32  +  J^T2 

=  -  2  In  (x  +  3)  +  2  In  (x  -  2)  +  In  (x  +  2)  +  C 

=  ini^±lH£z:^%c. 

EXERCISES 

Find  the  values  of  the  following  integrals : 

2.     I  r— ^ -dx.  5.     I  -r — ^ dx. 

r        x''-5x-}-5  r        x^-x-1 

^'  J  (x-lXx-2Xx-3f'''  ^'  J  (x-mx'-x-Qf''- 

76.  Table  of  integrals.  The  formulas  of  integration  used  in 
this  chapter  are  sufficient  for  the  solution  of  most  of  the  prob- 
lems which  occur  in  practice.  To  these  formulas  we  have  added 
a  few  others.  In  some  cases  they  represent  an  integral  which 
has  already  been  evaluated,  and  in  other  cases  they  are  the 
result  of  an  integration  by  parts.  In  all  cases  they  can  be 
verified  by  differentiating  both  sides  of  the  equation. 

These  collected  formulas  form  a  brief  table  of  integrals  which 
will  aid  in  the  solution  of  the  problems  in  this  book.  It  will  be 
noticed  that  some  of  the  formulas  express  the  given  integral  only 
in  terms  of  a  simpler  integral. 

I.  Fundamental 

1.  I  edu  =  c  I  du. 

2.  I  (du  -i-  dv  -^  dw  '  "')  =  I  du-i-  I  dv -\-  I  dw  •  ". 

3.  /  udv  =  uv  —  j  vdu. 


218  INTEGRATION 

II.  Algebraic 


4.  fu^du  =  ^^    (^^1) 

5.  f^  =  \nu. 
J    ^ 


^      ,      du         1 

D 


J  u^-\-a^      a  a  J 

r    du  \   ^    u  —  a     / 


7   r_jgi___^.i^^-/^ 

8.  f^'^i^^^du  =  1^2^  Va^-  2^-^+  «'sin-^^j. 

9.  fw  V  ^^:i^  ^i^  =  -  i  (a'  -  w') '• 

J  ^  +  2  ^  +  2^         (n  +  2^0) 

11.  I  =sm  '-. 

12.  I  •      =— Vg^— ?/« 

14.  I  y/u^±  a^du  =  \  [yr^u^  ±  d^  ±  a^ In {u  +-\/u^  ±  a^)\ 

15.  ^uVu""  ±aHu  =  \  (?/  ±  ay. 

J  71  +  2  71  +  2    J         (^  +  2^0) 

17.    r-^=  =  ln(^+V^7I7^). 


18. 

r    udu 

J  ^u^  ±  a^ 
r    u'^du 

=  ^u 

'±a\ 

^(n, 

-1), 
n 

%''  r  u^- 

19. 

-Wu'±a^ 

■''du 

J  ^u^±d' 

n 

''±a' 

(^n  ^  0) 


TABLE  219 


20.  I  — ,  =  -  sec^  -  • 

21.  r  V2  ai^  -u'du  =  ^\(u-  a)  V2  aw  -  u'  +  a'  sin-^  ^^^^1 


22.    /     .     ^^         =sin-^^~^ 


J  V2  aw  -  u"      '"        « 

III.  Trigonometric 

23.  I  sin  w  c?w  =  —  cos  w. 

24.  I  ^rn^udu  =  -  —  -  sin  2  w. 
J  2      4 

/I                   yi  —  \  r 
^m^udu  = sin"~^w  cos  u  H /  sin" ~^udu,     (n  ^  0) 
n                                n    J 

26.    j  cos  udu  =  sin  u. 

/u      1 
cos^udu  =  -  -f-  -  sin  2  u. 
2      4 

/I                      n  —  lr 
cos'^wc^M^ -cos"  ^wsinwH I  cos"~^wc?w.    Cw  =#=  0) 

29.  j  tan  udu  =  In  sec  w. 

30.  ftan^w^w  =^  ^^^^"     ^^  -  Ctmr-'^udu.    (n-l=^0} 

31.  j  (itnii  du  =  In  sin  u. 

32.  fctn^wc^w  =  -  ^^""    /^  -  Cctn''-''udu.    {n-l=^0^ 

33.  I  sec  udu  =  In  (sec  2*  +  tan  w). 

34.  I  sec^wc?w  =  tan  u. 

35.  I  CSC  wdfw  =  In  (esc  u  —  ctn  u). 


36.    /  csc^udu  —  —  ctn w. 


220  INTEGRATION 

37.  j  sec  u  tan  udu  =  sec  u. 

38.  I  CSC  u  ctn  udu  =  —  esc  u. 

/sin"*+^wcos""^w  ,   n  — 1  r  .   „  n-2    ^ 

sin"*i*cos'*wc?w= H —  I  sm'"i^  cos*  'wrfw. 

sin'"i^cos"wc?w= + — —      sm'"  ^uco^^'udu. 

m+n  m+nj         .      ,      ^  ^. 

(m  +  n4^  0) 

/sin"'+^wcos"+^w 
71+1 

^m  +  n  +  2  f^in^ucos^'  +  'udu.      (n  +  1^  0) 
n+1     J 


2. /si 


42.    I  sin'"w  cos"wc?w  = 


m  +1 

4-  ^?L±ZL±_^  rsin*»+^w  cos^wc^w.     (m  +1  ^  0) 
w  +  1     J 


IV.  Exponential 
43. /...«  =  . ».^ 

44.  /  a'^du  —  :, —  a". 
J  In  a 

,  ^      r       .    ,     ,        6;""  (a  sin  5w  —  J  cos  hu) 

45.  I  e««  sm  5w  c?i*  =  — ^^ ^-—T^ ^  • 

,  ^      r  ,     ,        e""  (a  cos  hu-\-h  sin  5w) 

J  a  +  0^ 

GENERAL  EXERCISES 

Find  the  values  of  the  following  integrals  : 


GENERAL  EXERCISES  221 


7.  f(2-\-e''-Ye''-dx.  24.    f— =^=  . 

8.  I     ,.  =dx.  25.    I  , 

r^!^  27    r ^^ 

/si„^(2x-l)cos'(2x-l)...  '^«-/(3,_,)V9x^-6.-: 

/   COS*-C?iC.  29.      I   7== 

J         5  J  (2ic  +  3)Vx=^+3a;-4 


10. 


11. 


12. 


13. 


/csc®4icft?ic.  30.     I      , 

14.  rsec^(a:-2)tan«(a!-2)rfa^.  31.    C—=£^= 
^  J   ^S  +  2x-x^ 

15.  I  Gtn(x —l)seG*(x —l)dx.  32      /  ^^ 

^  J   V7  + 

16.  /  csc^2£cctn'2a:c?ic.  33.    j  

17.  I  tan'3x"N/sec3a:c?a!.  34     j  

J  'J  4.' 

18.  /  ctn  2a:Vcsc  2iC6?iC.  35^    j  ! 

*J  1    O  X 


4  ic  —  4  ic^ 
dx 


-^-Qx-2x^ 
dx 


dx 


+  3 

10 


20. 
21. 
22 
23 


/CSC*  5  aj  "Vctn  5  x  dx.  o^     /^          ^^ 
^^- j  0:^-4^.4- 

J   sin*4x  ■  J  4ic-^+8ic+7 

r_csc^5^^^  38     f          ^^ 

J   •\/tan«5x      *  •  J  3a:2  4-4a^+l 

r    ^^    .  39  r     dx 

J   V25-4x^  ^^-j  90.2+ 5a; +1 


/dx  r  xdx 

V5372'  ^^-j^^  +  g 


222  INTEGRATION 

dx 


41. 


42. 


43. 


44. 


46. 
47 
48 
49 


51 


52 


57 


r  x^dx  gg    r 

J  3x«  +  7"  J   V9a:2- 24a:  4-14 

r    ^^    .  60.  r  ^^   . 

JxVx^-6  J  5^:^-4 


f    --^-         62.  r^^d 

J  V9-2x2  J  2x^- 

J   -Jx^-l  J 


5x 
dx 


5 


/dx  r           dx 

V4a:2+3*  'J4.x'-4.x 

r    dx  g^   r ^ 

j  V2^H^  J  25a:^-5a 


r_xdx__  r  dx 

•JV4^^^5*  ^«-j  3x^-4^-4 

/x^dx  r  dx 

.  r4^±L...  70.  r^ 


c?x 


+  3ic-2 


54.    I —;S=dx.  71.    /  (tan  3  a?  +  ctn  3  ic) Vic. 

/dx  r 


55 


cos2x 


J   V4ic*-^+4ic+7  *  J   ( 

r  dx  ^  r/sm2x  _  cos2_x\ 

J   V5x'-^+4cc— 1  'j\sinx         cosx  / 


GENERAL  EXEECISES  223 

7^'  J  sin'^dx.  91.    Cx5-+'dx. 


92.    /  x^e'^dx. 


J   14-  cos  4  ic  VA,,    I  3 

77      C         cos4x  /»             jp 

J  cos2a:-sin2x  ^3.    /  a^tan-i-^^a;. 

H dx.  ^^-    fx'smSxdx. 

X    ,       ^     X  J 

csc-  +  ctn- 

/»  96.    /  (ln2ic)V:r. 

79.  /  (sec^2a;-tan^2£c)^ic.  *^ 

r    , 96.    /  ln(3x4-V9a;2-4)c/a;. 

80.  /  Vl -h  sin  2  X  (^x.  »/ 

Oki       C  ^    ' 

81.  C-V^dx.                                      '  '  J  4:X^  + 

r  98  r^!±i^ 

\x-V7'dx.  J        x^- 

r  dx  99.  r  4.^+ 8.x -18 

84.  /  -.  J  8cf«H-12a^2-2a:-3 

r  101      r^^- 6x^-90. +  24, 

85.  /  a;^^^^^qr2^a:.  J       a:^  - 13  o;'^  +  36           * 

86.  r  ;^.  102.  r 

J  2a^«  +  l  Ji 


4x-3 
a^-18 


82. 


83. 


'^        c^x 


Vx2-9 


^      ^x 


V4  -  a;--^ 

89.  r^!^.  105.   f\^._%.^- 

90.  r ^^  106.    f\scHxdx. 

J  x'  y/x"  -  25  ^r. 


224  INTEGRATION 

\3in(x-^)seG(x-'^)dx.  115.    /  ^ 

\        Q/       \       Q/  J  4/-  x^/^T2 

Jin  Z  A  Jo 


4n 

108. 


JC  xdx 

J^'^Sgtan-l; 
I    ITS 


110 

111.      I         ^— ^^0^. 


113. 


: ax. 


/>!  117.  r 

109.     /     V^TT?7^^^  Ji 

JV2V(X2-1) 


'^  118.     r^_-^ 


119.    I     xsin~^cc(ix. 


r^  •  1 

/'^  __xdx__  ^^ 

2    -</2x  +  5*  121.    r 


120.    /   %^  sin  2  xc?a;. 


Vx'  +  l  122.     /    x^m^xdx. 

JC^       dx  r^ 

'    /  ,         '  123.    /    xHan-^xf^a;. 


CHAPTER  X 


APPLICATIONS 

77.  Review  problems.  The  methods  in  Chapter  III  for  de- 
termining areas,  volumes,  and  pressures  are  entirely  general, 
and  with  our  new  for- 
mulas of  integration  we  can 
now  apply  these  methods 
to  a  still  wider  range  of 
cases. 

Ex.  I.  Find  the  area  of  the 
ellipse -  +  ^  =  1. 

It  is  evident  from  the  sym- 
metry of  the  curve  (Fig.  86) 
that  one  fourth  of  the  required 
area  is  bounded  by  the  axis  of 
y,  the  axis  of  x,  and  the  curve. 

Constructing  the  rectangle  MNQP  as  the  element  of  area  dA,  we  have 


Fig.  86 


dA  =  ydx 

a 

Hence 

Jo  a 
2b 


Va^-x^dx. 


x'^dx 


—    a;  Va^—  x^  +  a-^sin-^- 
a  L  aJo 

=  irab. 

Ex.  2.  Find  the  area  bounded  by  the 
axis  of  X,  the  parabola  y^  =  kx,  and  the 
straight  line  y-\-2x  —  k  =  0  (Fig.  87). 

The  straight  line  and  the  parabola  intersect  at  the  point  ^'  (t  '  o) '  ^^^ 

/k      \  *  \4    2/ 

the  straight  line  intersects  OX  at  5  ( - ,  0  J .  Draw  CD  perpendicular  to  OX. 

If  we  construct  the  elements  of  area  as  in  Ex.  1,  they  will  be  of  different 

225 


226 


APPLICATIONS 


form  according  as  they  are  to  the  left  or  to  the  right  of  the  line  CD ;  for 
on  the  left  of  CD  we  shall  have 

dA  =  ydx  =  k'^x'^dx, 
and  on  the  right  of  CD  we  shall  have 

dA  =  ydx  =  (k  —  2x)  dx. 

It  will,  accordingly,  be  necessary  to  compute  separately  the  areas  ODC 

and  DBC  and  take  their  sum. 

k  * 

Area  ODC^Hkh^dx  =  1%^K^^  =  ^^^k\ 

k  ^ 

AreaDBC=f'\k-2x)dx=  hx  -  x-^V  =  ^\  k'^. 

I  4 

Hence  the  required  area  is  ^-g  P.    It  is  to  be  noted  that  the  area  DBC, 
since  it  is  that  of  a    right  triangle, 
could  have  been  found  by  the  formulas 
of  plane  geometry ;  for  the   altitude 

DC  =  -  and  the  base  DB  —  -  —  -  =  -, 
2  p       2      4      4 

and  hence  the  area  =  --  • 
lb 

Or  we  may  construct  the  element 
of  area  as  shown  in  Fig.  88. 

Then,  if  x^  and  x^  are  the  abscissas 
respectively  of  P^  and  P^, 


L2        4       3^0      48 


Hence 


k 


Ex.  3.  Let  the  ellipse  of  Ex.  1  be  represented  by  the  equations 

X  =  a  cos  ^,         y  =  h  sin  <^. 
Using  the  same  element  of  area,  and  expressing  y  and  dx  in  terms  of  ^, 

dA  =  {h  sin  <f))(—  a  sin  <lid<f>) 
=  —  ab  sin^  <f)  d<f>. 
As  X  varies  from  0  to  a,  <f>  varies  from  —  to  0 ; 


we  have 


hence 


A  =4:  C  ydx  =—  4  r  ab  sin^<f>d<f>. 


REVIEW  PROBLEMS 


227 


It  is  evident  from  formula  (1),  §  23,  that  the  sign  of  a  definite  integral 
is  changed  by  interchanging  the  limits.    Hence 


Jo 


Ex.  4.    Find  the  volume  of  the  ring  solid  generated  by  revolving  a 
circle  o^  radius  a  about  an  axis  in  its  plane  b  units  from  its  center  (b  >  a). 

Take  the  axis  of  revolution         Y 
as    OY  (Fig.  89)  and   the   line 
through  the  center  as  OX.   Then 
the  equation  of  the  circle  is 

(x  -  by  +f=  a2. 

A  straight  line  parallel  to  OX 

meets  the  circle  in  two  points  : 

P^,wh.ere  x=x^  =  b—Va'^—y'^,  and 

Pg,  where  x  =  X2=  b  +  Vd^  —  y'^. 

A  section  of  the  required  solid 

made  by  a  plane  through  P^Pg 

perpendicular  to  OF  is  bounded 

by  two  concentric   circles  with 

radii  MP^  =  x^   and  MP^  =  x^   respectively.    Hence,  if  d  V  denotes   the 

element  of  volume,  ,Tr      ,      o  ox  t 

dV=  (tt^I  —  TTx})  dy 


4  7r6Va2- 


y^dy. 


The  summation  extends  from  the  point  L,  where  y  =  —  a,  to  the  point  K, 
where  y  =^  a.  On  account  of  symmetry,  however,  we  may  take  twice  the 
integral  from  y  =  0  to  y  =  a.    Hence 

T- 

V=2  r"4  7r6  Va2. 
Jo 


yhly  =  2  7r'^a%. 


Ex.  5.  Find  the  pressure  on  a 
parabolic  segment,  with  base  2  b  and 
altitude  a,  submerged  so  that  its 
base  is  in  the  surface  of  the  liquid 
and  its  axis  is  vertical. 

Let  RQC  (Fig.  90)  be  the  parabolic  segment,  and  let  CB  be  drawn 
through  the  vertex  C  of  the  segment  perpendicular  to  PQ  in  the  surface  of 
the  liquid.  According  to  the  data,  RQ  —  2byCB  =  a.  Draw  LN  parallel 
to  TS,  and  on  LN  as  a  base  construct  an  element  of  area,  dA.   Let 

CM=x. 


228  APPLICATIONS 


Then 

dA  =(LN)dx. 

But,  from  §  30, 

ln'    cm 

whence 

a 

and  therefore 

dA=  —x^dx. 

The  depth  of  LN  below  the  surface  of  the  liquid  is  CB  —  CM  =  a  —  a; ; 
hence,  if  w  is  the  weight  of  a  unit  volume  of  the  liquid, 

dP  =  —  x*  (a  —  a:)  wdx, 
a* 

and  P  =  \    — r  ^  {a  —  x)  dx 

EXERCISES 

1.  Find  the  area  of  an  arch  of  the  curve  y  =  sin  x. 

a(  -        --\ 

2.  Find  the  area  bounded  by  the  catenary  y  =  -\e''  -\-  e  "/,  the 

axis  of  X,  and  the  lines  x  =  ^h. 

8  a' 

3.  Find  the  area  included  between  the  curve  y  =  -7, — 5  and 

X  -\-  4  a 
its  asymptote. 

4.  Find  the  area  of  one  of  the  closed  figures  bounded  by  the 
curves  1/^  =  16  cc  and  ?/^  =  x^. 

5.  Find  the  area  bounded  by  the  curve  7f=2{x  —  V)  and  the 
line  2x-3y  =  0. 

6.  Find  the  area  between  the  axis  of  x  and  one  arch  of  the 
cycloid  X  =  a(<^  —  sin  <^),  y  =  a(l  —  cos  </>). 

7.  Find  the  volume  of  the  solid  generated  by  revolving  about  OY 
the  plane  surface  bounded  by  OF  and  the  curve  x^  -{■  y^  =  a^. 

8.  Any  section  of  a  certain  solid  made  by  a  plane  perpendicular  to 
OX  is  an  isosceles  triangle  with  the  ends  of  its  base  resting  on  the 

x^        -»2 

ellipse  -^  4-  T^  =  1  and  its  altitude  equal  to  the  distance  of  the  plane 
from  the  center  of  the  ellipse.   Find  the  total  volume  of  the  solid. 

9.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  2  ?/  +  a  =  0  the  area  bounded  by  one  arch  of  the  curve  y  =  sin  x 
and  the  axis  of  x. 


REVIEW  PROBLEMS  229 

10.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  y  +  a  =  0  the  area  bounded  by  the  circle  x^  -\- 1/^=  c?. 

11.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  x  =  a  the  area  bounded  by  that  line  and  the  curve  ay^  =  x\ 

12.  A  right  circular  cone  with  vertical  angle  60°  has  its  vertex  at 
the  center  of  a  sphere  of  radius  a.  Find  the  volume  of  the  portion 
of  the  sphere  included  in  the  cone. 

13.  A  trough  2  ft.  deep  and  2  ft.  broad  at  the  top  has  semielliptical 
ends.    If  it  is  full  of  water,  find  the  pressure  on  one  end. 

14.  A  parabolic  segment  with  base  18  and  altitude  6  is  submerged 
so  that  its  base  is  horizontal,  its  axis  vertical,  and  its  vertex  in  the 
surface  of  the  liquid.    Find  the  total  pressure. 

15.  A  pond  of  15  ft.  depth  is  crossed  by  a  roadway  with  vertical 
sides.  A  culvert,  whose  cross  section  is  in  the  form  of  a  parabolic 
segment  with  horizontal  base  on  a  level  with  the  bottom  of  the 
pond,  runs  under  the  road.  Assuming  that  the  base  of  the  parabolic 
segment  is  4  ft.  and  its  altitude  is  3  ft.,  find  the  total  pressure  on 
the  bulkhead  which  temporarily  closes  the  culvert. 

16.  Find  the  pressure  on  a  board  whose  boundary  consists  of  a 
straight  line  and  one  arch  of  a  sine  curve,  submerged  so  that  the 
board  is  vertical  and  the  straight  line  is  in  the  surface  of  the  water. 

78.  Infinite  limits  or  integrand.  There  are  cases  in  which  it 
may  seem  to  be  necessary  to  use  infinity  for  one  or  both  of  the 
limits  of  a  definite  integral,  or  in  which  the  integrand  becomes 
infinite.  We  shall  restrict  the  discussion  of  these  cases  to  the 
solution  of  the  following  illustrative  examples : 

Ex.  1.  Find  the  area  bounded  by  the  curve  y  = —^  (Fig.  91),  the  axis  of 
X,  and  the  ordinate  x  =  1. 

It  is  seen  that  the  curve  has  the  axis  of  x  as  an  asymptote ;  and  hence, 
strictly  speaking,  the  required  area  is  not  completely 
bounded,  since  the  curve  and  its  asymptote  do  not 
intersect.  Accordingly,  in  Fig.  91,  let  0M=1  and 
ON  =  b(b>l)  and  draw  the  ordinates  MP  and 
NQ.   Then 

r^dx    r    n^         1 

area  MNQP  =  f   'if=-i=l-i. 
Ji    x^       L      a:Ji  0 

If  the  value  of  b  is  increased,  the  boundary  line  NQ  moves  to  the  right ; 
and  the  greater  b  becomes,  the  nearer  the  area  approaches  unity. 


k 


230 


APPLICATIONS 


We  may,  accordingly,  define  the  area  bounded  by  the  curve,  the  axis  of  x, 
and  the  ordinate  a:  =  1  as  the  limit  of  the  area  MNQP  as  b  increases  indefi- 
nitely, and  denote  it  by  the  symbol 


>^dx 


Ji     x^       t,^^Ji    x^ 
Ex.  2.  Find  the  area  bounded  by  the  curve  y  = 


(Fig.  92),  the 


/    2  i 

axis  of  X,  and  the  ordinates  x  =  0  and  x  =  a.  v  a  —  a;^ 

Since  the  line  x  =  a  is  an  asymptote  of  the  curve,  y  — >-  oo  when  x  — >►  a  ; 
furthermore,  the  area  is  not,  strictly  speaking,  bounded.   We  may,  however, 
find  the  area  bounded  on  the  right  by  the  ordinate 
X  =  a  —  h,  where  h  is  a  small  quantity,  with  the  result 

-h 


r<^-^       dx  r  •      1  xl"-^       .      ,  a  — 

I  — ==  =    sm-^  -  =  sin~^ 

Jo        Va^  —  x^      L  aJo  u 


If 


^     .      ,  a  —  li  .      n       TT 

0,  sm-i -^  sm-i  1  =  - 

a  2 


Hence  we  may  regard  —  as  the  value  of  the  area 


required,  and  express  it  by  the  integral 
dx  ^ .      r»-f^      dx 


/*«      a. 

Jo  ^/'^. 


Fig.  92 


dx  _^  .      r^  dx 
Vx      6  -»  ^•^i  Vx 


Va2  -  x^      h-.o'^o        Va^  x^ 

1    Vx 
Proceeding  as  in  Ex.  1,  we  place 

r 

But  f'^  =  \2VxT=2Vb-2, 

•^1    -y/x        L  Ji 

an  expression  which  increases  indefinitely  as  b  — >-  go  ;  hence  the  given 
integral  has  no  finite  value. 

We  accordingly  conclude  that  in  each  case  we  must  determine  a  limit, 
and  that  the  problem  has  no  solution  if  we  cannot  find  a  limit. 

79.  Area  in  polar  coordinates.  Let  0  (Fig.  93)  be  the  pole  and 
OM  the  initial  line  of  a  system  of  polar  coordinates  (r,  6),  OP^ 
and  OiJ  two  fixed  radius  vectors  for  which  0  =  6^  and  0  =  6^ 
respectively,  and  I^I^  any  curve  for  which  the  equation  is 
r=f(6).    Required  the  area  I^OI^. 


AREA  IN  POLAR  COORDINATES 


231 


To  construct  the  differential  of  area,  dA,  we  divide  the  angle 
J^OI^  into  parts,  d6.  Let  OF  and  0^  be  any  two  consecutive 
radius  vectors ;  then  the  angle  P0Q  =  d6.  With  0  as  a  center 
and  OP  as  a  radius,  we  draw  the  arc  of  a  circle,  intersecting  OQ 
at  E.  The  area  of  the  sector 
FOIt=^(OFyd6  =  ^r'd0. 

It  is  obvious  that  the  re- 
quired area  is  the  limit  of  the 
sum  of  the  sectors  as  their 
number  is  indefinitely  in- 
creased. Therefore  we  have 


dA  =  l7^de 


and 


A  = 


'^r'de. 


This  result  is  unchanged 
if  ij  coincides  with  0,  but 
in  that  case  01^  must  be  tangent  to  the  curve 
coincide  with  0. 


Fig.  93 


So  also  ij  may 


Ex.  1.  Find  the  area  of  one  loop  of  the  curve  r  =  a  sin  3  0  (Fig.  65,  §  51). 
As  the  loop  is  contained  between  the  two  tangents  ^  =  0  and  6  =  —,  the 


required  area  is  given  by  the  equation 


Jo 

=  ^p(i-cosQe)de 


12 

Ex.  2.  Find  the  area  bounded  by  the  lines  B=  —  -  and  6=  -t  the  curve 

^  4  4 

r  =  2  a  cos  $,  and  the  loop  of  the  curve  r  =  a  cos  2  0  which  is  bisected  by 
the  initial  line. 

Since  the  loop  of  the  curve  r  =^  a  cos  2  ^  is  tangent  to  the  line  OL 


(Fig.  94),  for  which  $  =  —  j,  and  the  line  ON,  for  which  ^  =-T,  it  is  evi- 


4' 


dent  that  the  required  area  can  be  found  by  obtaining  the  area  OLMNO, 
bounded  by  the  lines  OL  and  ON  and  the  curve  r  =  2  a  cos  6,  and  subtract- 
ing from  it  the  area  of  the  loop.  The  area  may  also  be  found  as  follows : 
Let  OP^Pr^  be  any  radius  vector  cutting  the  loop  r  =  a  cos  2  ^  at  Pj  and 
the  curve  r  =  2  a  cos  ^  at  P^.    Let  OP^  -  r^  and  OP^  =  r^.  Draw  the  radius 


232 


APPLICATIONS 


vector  OQ^Q^,  making  an  angle  dO  with  OP^P^.  With  OP^  and  OP^  as 
radii  and  0  as  a  center,  construct  arcs  of  circles  intersecting  OQ^Q^  at  R^ 
and  /?2  respectively.  Then  the  area  of  the  sector  P^OR^  is  \r^dd  and  the 
area  of  the  sector  P^OR^  is  I  r^dd.  We 
may  now  take  the  area  P^P^^R^R^  as  dA^ 
and  have 

dA  =  l{rl-rl)d6. 


Then    A=pl(^rl-rl)d6', 

~  i 

or,  since  the  required  area  is  symmetrical 
with  respect  to  the  line  OM,we  may  place 

A=:2nh(r.!-r^)de 
Jo. 


=!:<-'■' 


\')M. 


From  the  curve  r  =  2  a  cos  6,  we  have 


4  a^  cos^  6,  and  from  the  curve 


r  =  a  cos  2  $,  we  have  r^  =  a2cos2  2  6;  so  that  finally 
^=  r4^4a2cos2^-  a^cos^2e)d0 


'■•[ 


26+ sin  26-^- 


6      sin  4^-1* 


1 


(3  TT  +   8). 


EXERCISES 

1.  Find  the  total  area  of  the  lemniscate  r^=2a'^  cos  2  $. 

2.  Find  the  area  of  one  loop  of  the  curve  r  =  a  sin  ti  ^. 

3.  Find  the  total  area  of  the  cardioid  r  =  a(l  -f-  cos  ^). 

4.  Find  the  total  area  bounded  by  the  curve  r  =  5  +  3  cos  ^. 

5.  Find  the  area  of  the  loop  of  the  curve  r'^  =  a^  cos  2  ^  cos  ^ 
which  is  bisected  by  the  initial  line. 

6.  Find  the  area  bounded  by  the  curves  r  =  a  cos  3  ^  and  r  —  a. 

7.  Find  the  total  area  bounded  by  the  curve  r  =  3  +  2  cos  4  ^. 

Q 

8.  Find  the  area  bounded  by  the  curve  rcos^-=l  and  the 

77-  ^ 

lines  ^  =  0  and  6=  —  - 

9.  Find  the  area  bounded  by  the  curves  r  =  6  -f  4  cos  ^  and 
r  =  4  cos  Q. 

10.   Find  the  area  bounded  by  the  curves  r  =  a  cos  ^  and  r*=  a?-  cos  2  B. 


MEAN  VALUE 


233 


80.  Mean  value  of  a  function.    Letf(x)  be  any  function  of  x 
and  let  ?/  =f(x)  be  represented  by  the  curve  AB  (Fig.  95),  where 
OM=a  and  ON=h.    Take  the  points  Jf^,  Jf^'  *  * '»  ^«-i  so  as 
to    divide    distance    MN  into 
71   equal   parts,   each  equal  to 
dx^  and  at  the  points  Jf,  ilf^,  ilS/^, 
•  •  •,  iJ/l   ,  erect  the   ordinates 


^0'  y^  y^ 


yn -11    Then  the 


average,  or  mean,  value  of  these 
n  ordinates  is 


This  fraction  is  equal  to 

+yn-ddx 


Fig.  95 


{yo+yi+yi+" 

ndx 


y^dx-\-y^dx-^y^dx-\- 
h  —  a 


■^yn-idx 


If  n  is  indefinitely  increased,  this  expression  approaches  as  a 
limit  the  value 


1      r^ 

I    f(x)dx. 


This  is  evidently  the  mean  value  of  an  "  infinite  number  "  of 
values  of  the  function  f{x)  taken  at  equal  distances  between 
the  values  x  =  a  and  x  =  h.  It  is  called  the  mean  value  of  the 
function  for  that  interval. 

Graphically  this  value  is  the  altitude  of  a  rectangle  with  the 
base  MN  which  has  the  same  area  as  MNBA  which  equals 


f(x)dx. 


We  see  from  the  above  discussion  that  the  average  of  the 
function  y  depends  upon  the  variable  x  of  which  the  equal 
intervals  dx  were  taken,  and  we  say  that  the  function  was 
averaged  with  respect  to  x.  If  the  function  can  also  be  averaged 
with  respect  to  some  other  variable  which  is  divided  into  equal 
parts  the  result  may  be  different.  This  is  illustrated  in  the 
examples  which  follow. 


234  APPLICATIONS 

Ex.  1.  Find  the  mean  velocity  of  a  body  falling  from  rest  during  the 
time  t^  if  the  velocity  is  averaged  with  respect  to  the  time. 

Here  we  imagine  the  time  from  0  to  i?^  divided  into  equal  intervals  dt 
and  the  velocities  at  the  beginning  of  each  interval  averaged.  Proceeding 
as  in  the  text,  we  find,  since  v  =  gt,  that  the  mean  velocity  equals 


-±-j\,d<  =  iff,. 


Since  the  velocity  is  gt^  when  t  =  t^,  it  appears  that  in  this  case  the 
mean  velocity  is*  half  the  final  velocity. 

Ex.  2.  Find  the  mean  velocity  of  a  body  falling  from  rest  through  a 
distance  s^  if  the  velocity  is  averaged  with  respect  to  the  distance. 

Here  we  imagine  the  distance  from  0  to  s^  divided  into  equal  intervals 
(Is  and  the  velocities  at  the  beginning  of  each  interval  averaged.  Pro- 
ceeding as  in  the  text,  we  find,  since  v  =  V2  gs,  that  the  mean  velocity  is 


1       /"i. 
?i  -  0  Jo 


'2  OS  (Is  =  |V2 


gsas=  %^'^(JSy 


Since  the  velocity  is  V2  gs^,  when  s  =  s^,  we  see  that  in  this  case  the 
mean  velocity  is  two  thirds  the  final  velocity. 

EXERCISES 

1.  Find  the  mean  value  of  the  lengths  of  the  perpendiculars 
from  a  diameter  of  a  semicircle  to  the  circumference,  assuming  the 
perpendiculars  to  be  drawn  at  equal  distances  on  the  diameter. 

2.  Find  the  mean  length  of  the  perpendiculars  drawn  from  the 
circumferen-ce  of  a  semicircle  to  its  diameter,  assuming  the  perpen- 
diculars to  be  drawn  at  equal  distances  on  the  circumference. 

3.  Find  the  mean  value  of  the  ordinates  of  the  curve  y  =  sin  x 

TT 

between  x  =  0  and  x  =  —,  assuming  that  the  points  at  which  the 
ordinates  are  drawn  are  at  equal  distances  on  the  axis  of  x. 

4.  The  range  of  a  projectile  fired  with  an  initial  velocity  v^  and 

an  elevation  a  is  — ^  sin  2  a.    Find  the  mean  range  as  a  varies  from 

TT  .        ^. 

0  to  —  >  averaging  with  respect  to  a. 

5.  Find  the  mean  area  of  the  plane  sections  of  a  right  circular 
cone  of  altitude  h  and  radius  a  made  by  planes  perpendicular  to  the 
axis  at  equal  distances  apart. 


LENGTH  OF  PLANE  CURVE 


235 


6.  In  a  sphere  of  radius  a  a  series  of  right  circular  cones  is 
inscribed,  the  bases  of  which  are  perpendicular  to  a  given  diameter 
at  equidistant  points.    Find  the  mean  volume  of  the  cones. 

7.  The  angular  velocity  of  a  certain  revolving  wheel  varies  with 
the  time  until  at  the  end  of  5  min.  it  becomes  constant  and  equal  to 
200  revolutions  per  minute.  If  the  wheel  starts  from  rest,  what  is  its 
mean  angular  velocity  with  respect  to  the  time  during  the  interval 
in  which  the  angular  velocity  is  variable  ? 

8.  The  formula  connecting  the  pressure  p  in  pounds  per  square 
inch  and  the  volume  v  in  cubic  inches  of  a  certain  gas  is  jpv  =  20. 
Find  the  average  pressure  as  the  gas  expands  from  2\  cu.  in.  to  5  cu.  in. 

9.  Show  that  if  3/  is  a  linear  function  of  x,  the  mean  value  of  y 
with  respect  to  x  is  equal  to  one  half  the  sum  of  the  first  and  the 
last  value  of  y  in  the  interval  over  which  the  average  is  taken. 

81.  Length  of  a  plane  curve.  To  find  the  length  of  any 
curve  AB  (Fig.  96),  assume  n  —  1  points,  ij,  -^,  •  •  •,  -?_i,  be- 
tween A  and  B  and  connect  each  pair  of  consecutive  points  by 
a  straight  line.  The  length  of  AB  is 
then  defined  as  the  limit  of  the  sum 
of  the  lengths  of  the  n  chords  JiJ, 
^j^,  ^7J,  .  .  .,  J^_iB  as  n  is  increased 
without  limit  and  the  length  of  each 
chord  approaches  zero  as  a  limit.  By 
means  of  this  definition  we  have  already  - 
shown  (§§  39  and  52)  that 

in  Cartesian  coordinates,  and 


Jl 


Fig. 


(1) 


in  polar  coordinates. 
Hence  we  have 

and 


ds  =  Vdr^-hr^de'' 
s=  CVdx^-^df 
8  =  Cy/d7^-itr'dd\ 


(2) 
(3) 
(4) 


To  evaluate  either  (3)  or  (4)  we  must  express  one  of  the 
variables  involved  in  terms  of  the  other,  or  both  in  terms  of  a 
third.    The  limits  of  integration  may  then  be  determined. 


236  APPLICATIONS 

Ex.  1.  Find  the  length  of  the  parabola  -(p-  =  kx  from  the  vertex  to  the 
point  {a,  b). 

From  the  equation  of  the  parabola  we  find  2ydij  =  kdx.  Hence  formula 
(3)  becomes  either 


=XVn^-=X\'H^-' 


Either  integral  leads  to  the  result 


k,    2  ?>  +  V4  ?>2  +  y^2 


s  =  -f-V462_^P  + jln 

2k  4  k 

Ex.  2.    Find  the  length  of  one  arch  of  the  cycloid 

X  =  a(<}>  —  sin  <j>),  y  =  a(l—  cos  <f>). 

We  have  dx  =  a(l  —  cos  <f>)  d<fi,         dy  —  a  sin  <^ d<j> ; 

whence,  from  (1),     ds  —  a  V 2  —  2  cos  <fid<l>  =  2  a  sin  ~  6/^. 

J»27r  J, 

sin  ^d<j>  =  8  a. 

EXERCISES 

1.  Find  the  length  of  the  curve  3y^=(x  —  iy  from  its  point  of 
intersection  with  OX  to  the  point  (4,  3). 

2.  Find  the  length  of  the  catenary  y  =  ^Ve"  +  e  °/  from  x  =  0 
to  X  =  h. 

3.  Find  the  total  length  of  the  curve  x^-{-y^=  a*- 

4.  Find  the  total  length  of  the  curve  x  =  a  cos^<^,  y  =  cc  sin^<^. 

5.  Find  the  length  of  the  curve 

X  =  a  cos  <f>  -{-  a<f>  sin  <^,         y  =  a  sin  <l>  —  a<fi  cos  <^, 
from  (^  =  0  to  <^  =  4  TT. 

6.  Find  the  length  of  the  curve  x=  e~'cos  t,  y  =  e~'sin^,  between 

TT 

the  points  for  which  ^  =  0  and  t  =  —■ 

7.  Find  the  length  of  the  curve  r  =  a  cos*-  from  the  point  on 
the  curve  for  which  ^  =  0  to  the  pole. 

8.  Find  the  total  length  of  the  curve  r  =  a(l+  cos  0). 

9.  Show  that  the  length  of  the  logarithmic  spiral  r  =  e**^  between 
any  two  points  is  proportional  to  the  difference  of  the  radius  vectors 
of  the  points. 


WORK  237 

82.  Work.  By  definition  the  work  done  in  moving  a  body 
against  a  constant  force  is  equal  to  the  force  multiplied  by  the 
distance  through  which  the  body  is  moved.  If  the  foot  is  taken 
as  the  unit  of  distance  and  the  pound  is  taken  as  the  unit  of 
force,  the  unit  of  measure  of  work  is  called  2^  foot-pound.  Thus 
the  work  done  in  lifting  a  weight  of  25  lb.  through  a  distance 
of  50  ft.  is  1250  ft.-lb. 

Suppose  now  that  a  body  is  moved  along  OX  (Fig.  97)  from 
A(x  =  d)  to  B(x  =  6)  against  a  force  which  is  not  constant  but 
is  a  function  of  x^  expressed  by  f(x).    Let  the  line  AB  be 

divided  into  intervals  each  equal   — 1 1 m 1 — ^ 

^    ^         ^14-  f  4-1.        '\  ^    ^  ^^         B    ^ 

to  ax,  and  let  one  01  these  mter- 

vals  be  JfiV,  where  03f=  x.   Then 

the  force  at  the  point  M  is  f(x),  and  if  the  force  were  con- 
stantly equal  to  f(x^  throughout  the  interval  MN,  the  work 
done  in  moving  the  body  through  MN  would  hef(x)dx.  This 
expression  therefore  represents  approximately  the  work  actually 
done,  and  the  approximation  becomes  more  and  more  nearly 
exact  as  MN  is  taken  smaller  and  smaller.  The  work  done  in 
moving  from  ^  to  ^  is  the  limit  of  the  sum  of  the  terms  f(x)  dx 
computed  for  all  the  intervals  between  A  and  B,   Hence  we  have 

dW=f(x)dx 
and  W=  C  f(x)dx. 

Ex.  The  force  which  resists  the  stretching  of  a  spring  is  propor- 
tional to  the  amount  the  spring  has  been  already  stretched.  For  a  cer- 
tain spring  this  force  is  known  to  be  10  lb.  when  the  spring  has  been 
stretched  \  in.  Find  the  work  done  in  stretching  the  spring  1  in.  from 
its  natural  (unstretched)  length. 

If  F  is  the  force  required  to  stretch  the  spring  through  a  distance  ar, 
we  have,  from  the  statement  of  the  problem, 

F-kx] 

and  since  F  =  10  when  a:  =  ^,  we  have  k  =  20.   Therefore  JP  =  20  ar. 
Reasoning  as  in  the  text,  we  have 

W=  r^20x</2;  =  10in.-lb. 


/; 


238  APPLICATIONS 

EXERCISES 

1.  A  positive  charge  m  of  electricity  is  fixed  at  O.    The  repulsion 

Till 

on  a  unit  charge  at  a  distance  x  from  0  is  -^  •    Find  the  work  done 

in  bringing  a  unit  charge  from  infinity  to  a  distance  a  from  0. 

2.  Assuming  that  the  force  required  to  stretch  a  wire  from  the 

length  a  to  the  length  a  -\- x  i^  proportional  to  -  >  and  that  a  force 

of  1  lb.  stretches  a  certain  wire  36  in.  in  length  to  a  length  .03  in. 
greater,  find  the  work  done  in  stretching  that  wire  from  36  in.  to  40  in. 

3.  A  block  slides  along  a  straight  line  from  O  against  a  resistance 

equal  to  —^ :;'  where  k  and  a  are  constants  and  x  is  the  distance 

^  x^-{-  of 

of  the  block  from  0  at  any  time.    Find  the  work  done  in  moving  the 
block  from  a  distance  a  to  a  distance  a  Vs  from  0. 

4.  Find  the  foot-pounds  of  work  done  in  lifting  to  a  height  of 
20  ft.  above  the  top  of  a  tank  all  the  water  contained  in  a  full  cylin- 
drical tank  of  radius  2  ft.  and  altitude  10  ft. 

5.  A  bag  containing  originally  801b.  of  sand  is  lifted  through  a 
vertical  distance  of  8  ft.  If  the  sand  leaks  out  at  such  a  rate  that  while 
the  bag  is  being  lifted,  the  number  of  pounds  of  sand  lost  is  equal  to  a 
constant  times  the  square  of  the  number  of  feet  through  which  the  bag 
has  been  lifted,  and  a  total  of  20  lb.  of  sand  is  lost  during  the  lifting, 
find  the  number  of  foot-pounds  of  work  done  in  lifting  the  bag. 

6.  A  body  moves  in  a  straight  line  according  to  the  formula  x  —  cf^ 
where  x  is  the  distance  traversed  in  a  time  t.  If  the  resistance  of  the 
air  is  proportional  to  the  square  of  the  velocity,  find  the  work  done 
against  the  resistance  of  the  air  as  the  body  moves  from  ic  =  0  to  cc  =  a. 

7.  Assuming  that  above  the  surface  of  the  earth  the  force  of  the 
earth's  attraction  varies  inversely  as  the  square  of  the  distance  from 
the  earth's  center,  find  the  work  done  in  moving  a  weight  of  ic  pounds 
from  the  surface  of  the  earth  to  a  distance  a  miles  above  the  surface. 

8.  A  wire  carrying  an  electric  current  of  magnitude  C  is  bent 
into  a  circle  of  radius  a.  The  force  exerted  by  the  current  upon  a 
unit  magnetic  pole  at  a  distance  x  from  the  center  of  the  circle  in  a 
straight  line  perpendicular  to  the  plane  of  the  circle  is  known  to  be 

'  Find  the  work  done  in  bringing  a  unit  magnetic  pole  from 

infinity  to  the  center  of  the  circle  along  the  line  just  mentioned. 


GENERAL  EXEKCISES  239 

9.  A  piston  is  free  to  slide  in  a  cylinder  of  cross  section  S,  The 
force  acting  on  the  piston  is  pS,  where  p  is  the  pressure  of  the  gas 
in  the  cylinder,  and  is  7.71b.  per  square  inch  when  the  volume  v  is 
2.5  cu.  in.  Find  the  work  done  as  the  volume  changes  from  2  cu.  in.  to 
6  cu.  in.,  according  as  the  law  connecting  j^  and  v  is  (1)  pv  =  k  or 
(2)  pv'-^  =  k. 

GENERAL  EXERCISES 

1.  Find  the  area  of  the  sector  of  the  ellipse  4  cc^ -|- 9  y^  =  36 
cut  out  of  the  first  quadrant  by  the  axis  of  x  and  the  line  2y  =  x. 

2.  Find  the  area  of  each  of  the  two  parts  into  which  the  area  of 
the  circle  x^  -}-  y^  =  36  is  divided  by  the  curve  y^  =  x^. 

3.  Find  the  area  bounded  by  the  hyperbola  xy  =  12  and  the 
straight  line  x  +  y  —  S  =  0. 

4.  Find  the  area  bounded  by  the  parabola  x?—^ay  and  the 


0^2  +  4  a^ 

5.  Find  the  area  of  the  loop  of  the  curve  at/^  =  (a;  —  a)  (x  —  2  of. 

6.  Find  the  area  of  the  two  parts  into  which  the  loop  of  the 
curve  if-  =  x^  {^  —  X)  is  divided  by  the  line  x  —  y  =  ^. 

7.  Find  the  area  bounded  by  the  curve  xh/  -f-  a%'^  —  ahf  and  its 
asymptotes. 

8.  Find  the  area  bounded  by  the  curve  y'^ix^  +  a?)  =  a^x?  and  its 
asymptotes. 

9.  Find  the  area  bounded  by  the  curve  x  =  aco^6,y  =  h  sin^^. 

10.  Find  the  area  inclosed  by  the  curve  x  =  a  cos^^,  y  =  a  sin^^. 

1 1 .  Two  parabolas  have  a  common  vertex  and  a  common  axis,  but 
lie  in  perpendicular  planes.  An  ellipse  moves  with  its  plane  perpen- 
dicular to  the  axis  and  with  the  ends  of  its  axes  on  the  parabolas. 
Find  the  volume  generated  when  the  ellipse  has  moved  a  distance 
A  from  the  common  vertex  of  the  parabolas. 

12.  Find  the  volume  of  the  solid  formed  by  revolving  about  the  line 
x  =  ^L  the  figure  bounded  by  the  parabola  y'^  =  ^x  and  the  line  x  =  l. 

13.  A  right  circular  cylinder  of  radius  a  is  intersected  by  two 
planes,  the  first  of  which  is  perpendicular  to  the  axis  of  the  cylinder 
and  the  second  of  which  makes  an  angle  0  with  the  first.  Find  the 
volume  of  the  portion  of  the  cylinder  included  between  these  two 
planes  if  their  line  of  intersection  is  tangent  to  the  circle  cut  from 
the  cylinder  by  the  first  plane. 


240  APPLICATIONS 

14.  On  the  double  ordinate  of  the  curve  x'^  4- 1/3  =  a^  as  a  base, 
an  isosceles  triangle  is  constructed  with  its  altitude  equal  to  the 
ordinate  and  its  plane  perpendicular  to  the  plane  of  the  curve.  Find 
the  volume  generated  as  the  triangle  moves  from  x  =—  a  to  x  =  a. 

15.  Find  the  volume  of  the  solid  generated  by  revolving  about 

8  a' 
the  line  OY  the  figure  bounded  by  the  curve  i/  =    .^  ^  and  the 

line  y  =  (i' 

16.  Find  the  volume  of  the  solid  formed  by  revolving  about  the 
line  x  =  —  2  the  plane  area  bounded  by  that  line,  the  parabola  y^=zSXf 
and  the  lines  y  =  ±S. 

17.  Find  the  volume  formed  by  revolving  about  the  line  x  =  2 
the  plane  figure  bounded  by  the  curve  y^=  4:(2  —  x)  and  the  axis  of  y. 

18.  The  sections  of  a  solid  made  by  planes  perpendicular  to  OF 
are  circles  with  one  diameter  extending  from  the  curve  y^  =  4:x  to 
the  curve  y^=  A  —  4.x.  Find  the  volume  of  the  solid  between  the 
points  of  intersection  of  the  curves. 

19.  The  area  bounded  by  the  circle  x^  -{-  y^—  2ax  =  0  is  revolved 
about  OX,  forming  a  solid  sphere.  Find  the  volume  of  the  two  parts 
into  which  the  sphere  is  divided  by  the  surface  formed  by  revolving 

the  curve  ?/  =  7, about  OX. 

^       2a  —  X 

20.  Find  the  volume  of  the  two  solids  formed  by  revolving  about 
0  Y  the  areas  bounded  by  the  curves  x^  -\-  y^  =  5  and  y^=z  4:X. 

21.  Find  the  volume  of  the  solid  formed  by  revolving  about  OX 
the  area  bounded  by  OX,  the  lines  a;  =  0  and  x  =  a,  and  the  curve 

X 

y  =  X  -\-  ae^. 

22.  The  three  straight  lines  OA,  OB,  and  OC  determine  two  planes 
which  intersect  at  right  angles  in  OA.  The  angle  A  OB  is  45°  and  the 
angle  AOC  is  60°.  The  section  of  a  certain  solid  made  by  any  plane 
perpendicular  to  OA  is  a  quadrant  of  an  ellipse,  the  center  of  the 
ellipse  being  in  OA,  an  end  of  an  axis  of  the  ellipse  being  in  OB,  and 
an  end  of  the  other  axis  of  the  ellipse  being  in  OC.  Find  the  volume 
of  this  solid  between  the  point  0  and  a  plane  perpendicular  to  OA  at 
a  distance  of  two  units  from  0. 

23.  The  section  of  a  solid  made  by  any  plane  perpendicular  to  OX 
is  a  rectangle  of  dimensions  x'^  and  sin  x,  x  being  the  distance  of  the 
plane  from  0.  Find  the  volume  of  this  solid  included  between  the 
planes  for  which  cc  =  0  and  x  =  ir. 


GENERAL  EXERCISES  241 

24.  An  oil  tank  is  in  the  form  of  a  horizontal  cylinder  the  ends 
of  which  are  circles  4  ft.  in  diameter.  The  tank  is  full  of  oil,  which 
weighs  50  lb.  per  cubic  foot.  Calculate  the  pressure  on  one  end  of 
the  tank. 

25.  The  gasoline  tank  of  an  automobile  is  in  the  form  of  a  hori- 
zontal cylinder  the  ends  of  which  are  plane  ellipses  20  in.  high  and 
10  in.  broad.  Assuming  w  as  the  weight  of  a  cubic  inch  of  gasoline, 
find  the  pressure  on  one  end  of  the  tank  when  the  gasoline  is  15  in. 
deep. 

26.  A  horizontal  gutter  is  U-shaped,  a  semicircle  of  radius  3  in., 
surmounted  by  a  rectangle  6  in.  wide  by  4  in.  deep.  If  the  gutter  is 
full  of  water  and  a  board  is  placed  across  the  end,  how  much  pressure 
is  exerted  on  the  board  ? 

27.  The  end  of  a  horizontal  gutter  is  in  the  form  of  a  semicircle 
of  3  in.  radius,  the  diameter  of  the  semicircle  being  at  the  top  and 
horizontal.  The  gutter  receives  water  from  a  roof  50  ft,  above  the 
top  of  the  gutter.  If  the  pipe  leading  from  the  roof  to  the  gutter  is 
full,  what  is  the  pressure  on  a  board  closing  the  end  of  the  gutter  ? 

28.  A  circular  water  main  has  a  diameter  of  5  ft.  One  end  is  closed 
by  a  bulkhead,  and  the  other  is  connected  with  a  reservoir  in  which 
the  surface  of  the  water  is  20  ft.  above  the  center  of  the  bulkhead. 
Find  the  total  pressure  on  the  bulkhead. 

29.  Find  the  area  of  a  loop  of  the  curve  r^=  a^  sin  nO. 

30.  Find  the  area  swept  over  by  a  radius  vector  of  the  curve 

TT 

r  —  a  tan  ^  as  ^  changes  from  0  to  —  • 

4 

31.  Find  the  area  inclosed  by  the  curve  r  = ;:  and  the 

.  ^  1  —  cos  ^ 

4 

curve  r  = ;:  ♦ 

1  +  cos  ^ 

32.  Find  the  area  bounded  by  the  circles  r  =  a  cos  ^  and  r = a  sin  B. 

33.  Find  the  area  cut  off  from  one  loop  of  the  curve  r^  =  2  a^  sin  2  Q 
by  the  circle  r  —  a. 

34.  Find  the  area  of  the  segment  of  the  cardioid  r  =  a  (1  +  cos  ^) 
cut  off  by  a  straight  line  perpendicular  to  the  initial  line  at  a  dis- 
tance I  a  from  the  origin  0. 

35.  Find  the  area  cut  off  from  a  loop  of  the  curve  r  =  a  sin  3  ^  by 

the  circle  r  =  — - —  • 


242  APPLICATIONS 

36.  Find  the  area  cut  off  from  the  lemniscate  t^  =  2o?  cos  2  ^  by 

the  straight  line  r  cos  B  =  ' 

37.  Find  each  of  the  three  areas  bounded  by  the  curves  r  =  a 
and  r  =  a  (1  +  sin  &). 


38.  Find  the  mean  height  of  the  curve  y  =  ^^  ^   ^ — ^  between  the 


lines  X  =  —  2a  and  x  =  2a. 

39.  A  particle  describes  a  simple  harmonic  motion  defined  by  the 

during  a  complete  vibration  is  half  the  maximum  kinetic  energy  if 
the  average  is  taken  with  respect  to  the  time. 

40.  In  the  motion  defined  in  Ex.  39  what  will  be  the  ratio  of 
the  mean  kinetic  energy  during  a  complete  vibration  to  the  maxi- 
mum kinetic  energy,  if  the  average  is  taken  with  respect  to  the 
space  traversed  ? 

41.  A  quantity  of  steam  expands  according  to  the  \2jw  pv^-^  =  2000, 
p  being  the  pressure  in  pounds  absolute  per  square  foot.  Find  the 
average  pressure  as  the  volume  v  increases  from  1  cu.  ft.  to  5  cu.  ft. 

42.  Find  the  length  of  the  curve  y  —  a  In-^ ^  from  the  origin 

a  ^  —  "^ 

to  the  point  for  which  x  —  -- 

43.  Find  the  length  of  the  curve  y  =  In^ — r  between  the  points 
for  which  x  =1  and  x  =  2  respectively. 

44.  Find  the  total  length  of  the  curve  x  —  a  cos^<^,  y  =  h  sin^<^. 

Q 

45.  Find  the  total  length  of  the  curve  r  =  a  sin'-* 

o 

46.  Find  the  length  of  the  spiral  r  —  ad  from  the  pole  to  the  end 

of  the  first  revolution. 

k 

47.  If  a  center  of  force  attracts  with  a  magnitude  equal  to  — » 

where  x  is  the  distance  of  the  body  from  the  center,  how  much  work 
will  be  done  in  moving  the  body  in  a  straight  line  away  from  the 
center,  from  a  distance  a  to  a  distance  8  a  from  the  center  ? 

48.  A  body  is  moved  along  a  straight  line  toward  a  center  of 
force  which  repels  with  a  magnitude  equal  to  lex  when  the  body 
is  at  a  distance  x  from  the  center.  How  much  work  will  be  done 
in  moving  the  body  from  a  distance  2  a  to  a  distance  a  from  the 
center  ? 


GENERAL  EXERCISES  243 

49.  A  central  force  attracts  a  body  at  a  distance  x  from  the  center 

k 
by  an  amount  —  •   Find  the  work  done  in  moving  the  body  directly 

away  from  the  center  from  a  distance  a  to  the  distance  2  a. 

50.  How  much  work  is  done  against  hydrostatic  pressure  in  rais- 
ing a  plate  2  ft.  square  from  a  depth  of  20  ft.  to  the  surface  of  the 
water,  if  it  is  kept  at  all  times  parallel  to  the  surface  of  the  water  ? 

51.  A  spherical  bag  of  radius  5  in.  contains  gas  at  a  pressure 
equal  to  15  lb.  per  square  inch.  Assuming  that  the  pressure  is  in- 
versely proportional  to  the  volume  occupied  by  the  gas,  find  the 
work  required  to  compress  the  bag  into  a  sphere  of  radius  4  in. 


CHAPTER  XI 
REPEATED  INTEGRATION 
83.  Double  integrals.    The  symbol 

I     j    f(x,y)dxdy,  (1) 

in  which  a  and  h  are  constants  and  y^  and  y^  are  either  con- 
stants or  functions  of  x^  indicates  that  two  integrations  are  to 
be  carried  out  in  succession.   The  first  integral  to  be  evaluated  is 

f(x,  y)  dxdy, 

where  x  and  dx  are  to  be  held  constant.  The  result  is  a  func- 
tion of  x  only,  multiplied  by  dx;  let  us  say,  for  convenience, 
F(x)dx. 

The  second  integral  to  be  evaluated  is,  then, 


X 


b 

F(x)dx, 


which  is  of  the  familiar  type. 
Similarly,  the  symbol 

f(x,y)dydx,  (2) 


Xi 


where  a  and  h  are  constants  and  x^  and  x^  are  either  constants 
or  functions  of  y,  indicates  first  the  integration 


X 


f(x,  y)  dydx. 


in  which  y  and  dy  are  handled  as  constants,  and  afterwards 
integration  with  respect  to  y  between  the  limits  a  and  5. 

244 


DOUBLE  INTEGRALS         '  245 


3    /»2 


Ex.  1.    Evaluate  j     C  xydxdy. 

The  first  integral  is 

I    xy dx dy  =  \\ xy'^ dx\   =2xdx. 

The  second  integration  is 

f%xdx=  [x2]'  =  9. 

Ex.2.    Evaluate  f    C  ""  {x^  ^^  y'^)dxdy. 

Jo   J\—x 

The  first  integration  is 

f~    (a:2  +  y'^)dxdy  =  [(x'^y  +  i/)^ix]J-f  =  (:r  -  2  x^  +  ^x^-^x^)dx. 

The  second  integration  is 

f\x-2x^-^  ^x^-  \x^)dx=  ^V 
Jo 

2a        i^ 

Ex.  3.   Evaluate  f  "  [^''y^dydx. 

The  first  integration  is 

fj^y^dydx=:  \y'xdyY~^=^dy. 
The  second  integration  is 

Jo    4a   ^       120  aJo        5 

A  definite  integral  in  one  variable  has  been  shown  to  be  the 
limit  of  a  sum,  from  which  we  infer  that  formula  (1)  involves 
first  the  determination  of  the  limit  of  a  sum  with  respect  to  y, 
followed  by  the  determination  of  the  limit  of  a  sum  with  respect 
to  X.  The  application  of  the  double  integral  comes  from  its 
interpretation  as  the  limit  of  a  double  summation. 

How  such  forms  arise  in  practice  will  be  illustrated  in  the 

following  sections. 

EXERCISES 

Find  the  values  of  the  following  integrals : 

1.  r     r^^dydx.  3.     j      yx'dydx. 

2.  I      {''xydxdy.  4,     j      j     cos -dxdy. 


246 


EEPEATED  INTEGRATION 


8. 


n^'    dydx 
J    [     \x^ ■\- f)dxdy. 
0     Jq 

n 

Jo    Jo 


.a(l  +  co8  0) 


'dOdr. 


a  cos  '2  6 


rdOdr 


9 
10 
11 
12 


.^     ^iaBind 


r^dOdr. 


Jo     Jo 

■n. 

3 

rl  r«         dOdr 

4 

ir 

Jo     Jo 


r  sin  Odd  dr. 


COB  2d 

,!r      r,V2  8in2( 


r  cos  Odd  dr. 


84.  Area  as  a  double  integral.  Let  it  be  required  to  find  an 
area  (such  as  is  shown  in  Fig.  98)  bounded  by  two  curves,  with 
the  equations  ^i=/i(^)  and  y^-=f^ix)  intersecting  in  points  for 


Fig.  98 


which  x—a  and  x—h  respectively.  Let  the  plane  be  divided  into 
rectangles  by  straight  lines  parallel  to  OX  and  0  Y  respectively. 
Then  the  area  of  one  such  rectangle  is 


dA=  dxdy, 


(1) 


where  dx  is  the  distance  between  two  consecutive  lines  parallel 
to  OY^  and  where  dy  is  the  distance  between  two  consecutive 
lines  parallel  to  OX.   The  sum  of  the  rectangles  which  are  either 


AREA  AS  DOUBLE  INTEGRAL  247 

wholly  or  partially  within  the  required  area  will  be  an  approx- 
imation to  the  required  area,  but  only  an  approximation,  because 
the  rectangles  will  extend  partially  outside  the  area.  We  assume 
as  evident,  however,  that  the  sum  thus  found  becomes  more 
nearly  equal  to  the  required  area  as  the  number  of  rectangles 
becomes  larger  and  dx  and  dy  smaller.  Hence  we  say  that  the 
required  area  is  the  limit  of  the  sum  of  the  terms  dxdy. 

The  summation  must  be  so  carried  out  as  to  include  every 
rectangle  once  and  only  once.  To  do  this  systematically  we 
begin  with  any  rectangle  in  the  interior,  such  as  PQRS^  and  add 
first  those  rectangles  which  lie  in  the  vertical  column  with  it. 
That  is,  we  take  the  limit  of  the  sum  of  dxdy,  with  x  and  dx 
constant  and  y  varying  from  y^=zf^(x)  to  y^=f^{^-  This  is 
indicated  by  the  symbol 

rdxdy  =  iy-y^dx=^U^i^~)-fSx)-\Ax.  (2) 

This  is  the  area  of  the  strip  TXJVW.  We  are  now  to  take 
the  limit  of  the  sum  of  all  such  strips  as  dx  approaches  zero 
and  X  varies  from  a  to  h. 

We  have  then 

^=  r(%-y,)''^=  r'[/.(^)-/.(^)]<?^-     (3) 

%J  a  U  a 

If  we  put  together  what  we  have  done,  we  see  that  we  have 
^=  r    r^dxdy,  (4) 

This  discussion  enables  us  to  express  the  area  as  a  double 
integral.  It  does  not,  however,  give  us  any  more  convenient  way 
to  compute  the  area  than  that  found  in  Chapter  III,  for  the  result 
(2)  is  simply  what  we  may  write  down  at  once  for  the  area  of  a 
vertical  strip  (see  Ex.  3,  §  23). 

If  it  should  be  more  convenient  first  to  find  the  area  of  a 
horizontal  strip,  we  may  write 


-a> 


dx,  (5) 


248 


REPEATED  INTEGRATION 


Consider  a  similar  problem  in  polar  coordinates.  Let  an 
area,  as  in  Fig.  99,  be  bounded  by  two  curves  r^=f^{6) 
and  r^=f^(0^,  and  let  the  values  of  6  corresponding  to  the 
points  B  and  C  be  6^  and  6^  respectively.  The  plane  may  be 
divided  into  four-sided  figures  by  circles  with  centers  at  0  and 
straight  lines  radiating 
from  0.  Let  the  angle 
between  two  consecu- 
tive radii  be  dO  and 
the  distance  between 
two  consecutive  circles 
be  d7\  We  want  first 
the  area  of  one  of  the 
quadrilaterals  such  as 
PQRS.  Here  OF  =  r, 
PQ  =  dr,  and  the  angle 
FOS=de.  By  geom- 
etry the  area  of  the 
sector  FOS=  i  r^  dd  and 
the  area  of  the  sector 
QOR  =  l(r-^drydd', 

therefore  FQRS  =  J  (^  +  t^r) Vl9  -  \  r^dO  =  rdrdd  +  i  (drydd. 
Now  as  dr  and  dO  approach  zero  as  a  limit  the  ratio  of  the  second 
term  in  this  expression  to  the  first  term  also  approaches  zero, 
since  this  ratio  involves  the  factor  dr.  It  may  be  shown  that  the 
second  term  does  not  affect  the  limit  of  the  sum  of  the  expression, 
and  we  are  therefore  justified  in  writing  as  the  differential  of  area 

dA  =  rdedr,  (6) 

The  required  area  is  the  limit  of  the  sum  of  these  differ- 
entials. To  find  it  we  first  take  the  limit  of  the  sum  of  the 
quadrilaterals,  such  as  FQBS,  which  lie  in  the  same  sector  UOV. 
That  is,  we  integrate  rdddr,  holding  6  and  dO  constant  and  al- 
lowing r  to  vary  from  r^  to  r^.    We  have 


£ 


rdedr  =  \(rl-rl)de, 
which  is  the  area  of  the  strip  TUVW, 


(7) 


CENTER  OF  GRAVITY  249 

Finally  we  take  the  limit  of  the  sum  of  the  areas  of  all  such 
strips  in  the  required  area  and  have 

A=f\cr!-r^)de.  (8) 

If  we  put  together  what  we  have  done,  we  may  write 

rdOdr,  (9) 


J  9,     Jr. 


It  is  clear  that  this  formula  leads  to  nothing  which  has  not 
been  obtained  in  §  79,  but  it  is  convenient  sometimes  to  have 
the  expression  (9). 

85.  Center  of  gravity.  It  is  shown  in  mechanics  that  the  cen- 
ter of  gravity  of  n  particles  of  masses  m^^  w^,  •  •  •,  w„  lying  in  a 
plane  at  points  whose  coordinates  are  (x^^  y^^  (x^^  y^^  . .  .,  {x^^  y^ 
respectively  is  given  by  the  formulas 

-  ^  m^x^-\-m^x^^  .  .  .  +  m^x^  ^ 

m^y^Jrm^^^r"'^m^y^ 
m^-\-  m^-\-  '  '  '  ■\-  m^ 

This  is  the  point  through  which  the  resultant  of  the  weights 
of  the  particles  always  passes,  no  matter  how  the  particles  are 
placed  with  respect  to  the  direction  of  the  earth's  attraction. 

We  now  wish  to  extend  formulas  (1)  so  that  they  may  be 
applied  to  physical  bodies  in  which  the  number  of  particles  may 
be  said  to  be  infinite.  For  that  purpose  we  divide  the  body 
into  n  elementary  portions  such  that  the  mass  of  each  may  be 
considered  as  concentrated  at  a  point  (x^  y).  Then,  if  m  is  the 
total  mass  of  the  body,  the  mass  of  each  element  is  dm.  We  have 
then  to  replace  the  m's  of  formula  (1)  by  dm  and  to  take  the 
limit  of  the  sums  involved  in  (1)  as  the  number  n  is  indefinitely 
increased  and  the  elements  of  mass  become  indefinitely  small. 
There  result  the  general  formulas 

I  xdm  I  ydm 

-^  =  1- yJ (2) 

\  dm  I  dm 


250 


REPEATED  INTEGRATION 


To  apply  these  formulas  we  consider  first  a  slender  wire  so 
fine  and  so  placed  that  it  may  be  represented  by  a  plane  curve. 
More  strictly  speaking,  the  curve  may  be  taken  as  the  mathe- 
matical line  which  runs  through  the  center  of  the  physical  wire. 
Let  the  curve  be  divided  into  elements  of  length  ds.  Then, 
if  c  is  the  area  of  the  cross  section  of  the  wire  and  D  is  its 
density,  the  mass  of  an  element  of  the  wire  is  Dcds,  For  con- 
venience we  place  Dc  =  p  and  write 

dm  =  p  ds, 

where  /9  is  a  constant.  If  this  is  substituted  in  (2),  the  constant 
p  may  be  taken  out  of  the  integrals  and  canceled,  and  the 
result  may  be  written  in  the  form 


sx 


=  j  xds,         s?/  =  I  i/ds, 


(3) 


where  s  on  the  left  of  the  equations  is  the  total  length  of  the 
curve.   These  formulas  give  the  center  of  gravity  of  a  plane  curve. 

Ex.  1.    Find  the  center  of  gravity  of  one  fourth  of  the  circumference 
of  a  circle  of  radius  a. 

Here  we  know  that  the  total  length  is  |  Tra ;  so  that,  from  (3),  we  have 

Y 


^  irax  =  I  xds,         \  iray  —  (  yds. 

To  integrate,  it  is  convenient  to  in- 
troduce the  central  angle  <f>  (Fig.  100) ; 
whence  x  =  a  cos  <f),  y  =  a  sin  <f>,  ds  =  ad<}>. 

■JT 

Then    I  irax  z=  C^a^  cos  ^  d<l>, 
Jo 


\iray 


=  /o 


whence 


2 
0 

2a 


a^  sin  ^  (/</> ; 
_      2a 

y  =  — 


Fig.  100 


Consider  next  a  thin  plate,  which  may  be  represented  by  a 

plane   area.     Strictly   speaking,   the   area  is   that  of  a  section 

through  the  middle  of  the  plate.    If  t  is  the  thickness  of  the 

plate  and  D  its  density,  the  mass  of  an  element  of  the  plate 

with  the  area  dA  is  Dt  dA.    For  convenience  we  place  Dt  =  p 

and  write  ,  ,  , 

dm  =  pa  A, 


CENTER  OP  GRAVITY 


251 


where  /o  is  a  constant.    If  this  is  substituted  in  (2)  and  the  p's 
are  canceled,  we  have 


Ax 


=  /  xdA,         Ap=  I  ydA, 


(4) 


where  A  is  the  total  area.    These  formulas  give  the  center  of 
gravity  of  a  plane  area. 

Ex.  2.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 
^2  =  kx,  the  axis  of  x,  and  the  ordinate  through  the  point  (a,  b)  of  the 
parabola  (Fig.  101). 

We  place  dA  =  dxdy  in  (4)  and  have 

A  X  =  ffx  dx  dy,         A  y  =  ffy  dx  dy. 

To  evaluate,  we  choose  the  element  dxdy 
inside  the  area  in  a  general  position,  and  first 
sum  with  respect  to  y  along  a  vertical  strip. 
We  shall  denote  by  y^  the  value  of  y  on  the 
parabola,  to  distinguish  it  from  the  general 
values  of  y  inside  the  area.  The  first  integra- 
tion gives  us,  therefore,  respectively 


Fig.  101 


r  ^xdxdy  —  xy^dx     and       j    ^ ydxdy  —  \y\  dx, 


so  that  we  have  Ax  =  Cxy^dx,         Ay  =  fhVi 


dx. 


On  examination  of  these  results  we  see  that  each  contains  the  factor 
y^dx  (which  is  the  area  (§  22)  of  an  elementary  vertical  strip),  multiplied 
respectively  by  x  and  \  y^,  which  are  the  coordinates  of  the  middle  point  of 
the  ordinate  y^  These  results  are  the  same  as  if  we  had  taken  dA  =  y^  dx  in 
the  general  formula  (4),  and  had  taken  the  point  (x,  y)  at  which  the  mass 
of  dA  is  concentrated  as  (x,  ly^),  which  is  in  the  limit  the  middle  point  of 
dA.  In  fact  this  is  often  done  in  computing  centers  of  gravity  of  plane 
areas,  and  the  first  integration  is  thus  avoided. 

Now,  from  the  equation  of  the  parabola  y^  =  kx,  and  to  complete  the 
integration,  we  have  to  substitute  this  value  for  y^  and  integrate  with 
respect  to  x  from  x  =  0  to  x  =  a.    We  have 

Ax=pk^x^dx=  I  ArM,         Ay=P  \kxdx  =  ^- 

&2  2 

But,  from  the  equation  of  the  curve,  k=  —  and,  by  §  23,  J  =  -  ah. 

a  o 


252 


REPEATED  INTEGRATION 


Substituting  these  values  and  reducing,  we  have  finally 

In  solving  this  problem  we  have  carried  out  the  successive  integrations 
Separately,  in  order  to  show  clearly  just  what  has  been  done.  If  now  we 
collect  all  this  into  a  double  integral,  we  have 

/»  a    /.  •\/kj:  P^   n  "^kx 

Ax  =  I      I        xdxmj.         Aij  =  \      |         ydxdy. 
Jo  Jo  Jo  Jo 

Ex.  3.    Find  the  center  of  gravity  of  a  sextant  of  a  circle  of  radius  a. 

To  solve  this  problem  it  is  convenient  to  place  the  sextant  so  that  the 
axis  of  X  bisects  it  (Fig.  102)  and  to  use 
polar  coordinates. 

From  the  symmetry  of  the  figure  the 
center  of  gravity  lies  on  OX,  so  that  we 
may  write  at  once  y  =  0.  To  find  x  take 
an  element  of  area  rdOdr  in  polar  coor- 
dinates and  place  a:  =  r  cos  B.  We  have 
then,  from  (4), 

TT 

Ax=   f^  fr^  cos  SdOdr, 

J    n  Jo 

where  A  =  }  ira^,  one  sixth  the  area  of  a 
circle.  In  the  first  integration  6  and  dO 
are  constant,  and  the  summation  takes 
place  along  a  line  radiating  from  0  with 

r  varying  from  0  to  a.   The  angle  6  then  varies  from  ~  ^  to  ^  '  ^^^  thns  the 
entire  area  is  covered.   The  solution  is  as  follows : 


Fig.  102 


ira^x 


f[ 


la^cosede=^a^-y 


whence 


2a 


Consider  now  a  solid  of  revolution  formed  by  revolving  the 
plane  area  (Fig.  103)  ABCD  about  OF  as  an  axis.  It  is  assumed 
that  the  equation  of  the  curve  CD  is  given.  It  is  evident  from 
symmetry  that  the  center  of  gravity  of  the  solid  lies  on  OF,  so 
that  we  have  to  find  only  y. 

Let  ^Fbe  any  element  of  volume.  Then  dm  =  pdV,  where  p  is 
the  density  and  is  assumed  constant.  Substituting  in  (2),  we  have 


Vy 


=jya 


(5) 


CENTER  OF  GRAVITY 


253 


Let  the  solid  be  divided  into  thin  slices  perpendicular  to  OF, 
as  was  done  in  §  26,  and  let  the  summation  first  take  place  over 
one  of  these  slices.  In  this  summa- 
tion y  is  constant,  and  the  result 
of  the  summation  is  therefore  y 
times  the  volume  of  the  slice.  It  is 
therefore  y(jrx^dy').  We  have  now 
to  extend  the  summation  over  all 
the  slices.    This  gives  the  result 


Vy 


-i: 


irx^ydy, 


where  0A  =  a  and  OB  =  h. 

It  is  to  be  noticed  that  this  result 
is  what  we  obtain  if  we  interpret 

dm  in  (2)  as  the  mass  of  the  slice  and  consider  it  concentrated 
at  the  middle  point  of  one  base  of  the  slice. 

Ex.  4.    Find  the  center  of  gravity  of  a  right  circular  cone  of  altitude  h 
and  radius  a  (Fig.  104). 

This  is  a  solid  of  revolution  formed  by  revolv- 
ing a  right  triangle  about  OY.  However,  the 
equation  of  a  straight  line  need  not  be  used,  as 

similar  triangles  are  simpler.    We  have  -  =  - ; 

a  y    ^ 

vv'hence  x  =  -  y.    The  volume  V  is  known  to  be 
\Tra%.   Therefore,  from  (6),  we  have 

whence  y  =  1 6.  Pig.  104 

EXERCISES 

1.  Show  that  the  center  of  gravity  of  a  semicircumference  of 
radius  a  lies  at  a  distance  of  —  from  the  center  of  the  circle  on 

TT 

the  radius  which  bisects  the  semicircumference. 

2.  Show  that  the  center  of  gravity  of  a  circular  arc  which  subtends 

an  ansrle  a  at  the  center  of  a  circle  of  radius  a  lies  at  a  distance  —  sin  - 
^  a         2 

from  the  center  of  the  circle  on  the  radius  which  bisects  the  arc. 


1 


254  REPEATED  INTEGRATION 

n       —  —  — 

3.  A  wire  hangs  so  as  to  form  the  catenary  ?/  =  -  (e"  -h  e  "). 

Find  the  center  of  gravity  of  the  piece  of  the  curve  between  the 
points  for  which  x  =  0  and  x  =  a. 

4.  Find  the  center  of  gravity  of  the  arc  of  the  cycloid 
X  ==  a(<l>  —  sin  <^),  y  =  a(l—  cos  <^),  between  the  first  two  sharp 
points. 

5.  Find  the  center  of  gravity  of  a  parabolic  segment  of  base  2  b 
and  altitude  a. 

6.  Find  the  center  of  gravity  of  a  quadrant  of  the  area  of  a  circle. 

7.  Find  the  center  of  gravity  of  a  triangle. 

8.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
1/  =  sin  X  and  the  axis  of  x  between  a;  =  0  and  x  =  tt. 

9.  Find  the  center  of  gravity  of  the  plane  area  bounded  by  the 
two  parabolas  y^=  20  ic  and  x^  =  20  y. 

10.  Find  the  center  of  gravity  of  a  figure  which  is  composed  of 
a  rectangle  of  base  2  a  and  altitude  b  surmounted  by  a  semicircle 
of  radius  a. 

11.  Find  the  center  of  gravity  of  the  area  bounded  by  the  first 
arch  of  the  cycloid  (Ex.  4)  and  the  axis  of  x. 

12.  Show  that  the  center  of  gravity  of  a  sector  of  a  circle  lies  at 
a  distance  -^  sin  -  from  the  vertex  of  the  sector  on  a  line  bisecting 
the  angle  of  the  sector,  where  a  is  the  angle  and  a  the  radius. 

13.  Find  the  center  of  gravity  of  the  area  bounded  by  the  cardioid 
r  =  a(l-{-  cos  6). 

14.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
r  =  2  cos  ^  +  3. 

15.  Find  the  center  of  gravity  of  a  solid  hemisphere. 

16.  Find  the  center  of  gravity  of  a  solid  formed  by  revolving 
about  its  altitude  a  parabolic  segment  of  base  2  b  and  altitude  a. 

17.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 
about  OY  the  plane  figure  bounded  by  the  parabola  y^  =  kx,  the  axis 
of  y,  and  the  line  y  —  k. 

18.  Find  the  center  of  gravity  of  the  solid  bounded  by  the  sur- 
faces of  a  right  circular  cone  and  a  hemisphere  of  radius  a,  with  the 
base  of  the  cone  coinciding  with  the  base  of  the  hemisphere  and  the 
vertex  of  the  cone  in  the  surface  of  the  hemisphere. 


CENTER  OF  GRAVITY  256 

86.  Center  of  gravity  of  a  composite  area.  In  finding  the  cen- 
ter of  gravity  of  a  body  which  is  composed  of  several  parts  the 
following  theorem  is  useful: 

If  a  body  of  mass  M  is  composed  of  several  parts  of  masses 
J[fj,  iHfg,  •  •  'i  M^^  and  if  the  centers  of  gravity  of  these  parts  are 
respectively  (x^,  y^,  (x^,  y^,  -  -  .,(^„,  yj,  then  the  center  of  grav- 
ity of  the  composite  body  is  given  by  the  formulas 

Mi  =  M^\  +  M^^+  .  .  .  +  Jf„^„, 

My  =  M^y^+M^y^+...  +  M^^.  ^^ 

We  shall  prove  the  theorem  for  the  x  coordinate.  The  proof 
for  y  is  the  same. 

By  §  85  we  have,  for  the  original  body, 


Mx 


=  I  xdm,  (2) 


where  the  integration  is  to  be  taken  over  all  the  partial  masses 
J/j,  M^,  •  •  •,  Jf„  into  which  the  body  is  divided.    But  we  have  also 


=  jxjm^,  (3) 


^2^2 


/' 


i£ic„  =  /  x_  dm^, 


where  the  subscripts  indicate  that  the  integration  in  each  case 

is  restricted  to  one  of  the  several  bodies.    But  formula  (2)  can 

be  written  r  r  r 

Mx  =  I  x^dm^-\-  I  x^dm^-^ \-  j  x^dm^ ; 

and,  by  substitution  from  (3),  the  theorem  is  proved. 

Ex.  Find  the  center  of  gravity  of  an  area  bounded  by  two  circles  one 
of  which  is  completely  inside  the  other. 

Let  the  two  circles  be  placed  as  in  Fig.  105,  where  the  center  of  the 
larger  circle  of  radius  a  is  at  the  origin,  and  the  center  of  the  smaller 
circle  of  radius  b  is  on  the  axis  of  a:  at  a  distance  c  from  the  origin. 


W 


256 


REPEATED  IKTEGRATION 


The  area  which  can  be  considered  as  composed  of  two  parts  is  that  of 
the  larger  circle,  the  two  parts  being,  first,  the  smaller  circle  and,  second,  the 
irregular  ring  whose  center  of  gravity 
we  wish  to  find.  Now  the  center  of 
gravity  of  a  circle  is  known  to  be  at 
its  center.  Therefore,  in  the  formula 
of  the  theorem,  we  know  (x,  y),  which 
is  on  the  left  of  the  equation,  to  be 
(0, 0),  and  (x^,  y^)  to  be  (c,  0),  and  wish 
to  find  (Xg,  y^)- 

Since  we  are  dealing  with  areas, 
we  take  the  masses  to  be  equal  to  the 
areas,  and  have,  accordingly,  M  =  tto^ 
(the  mass  of  the  larger  circle),  M^=  nh^ 
(the  mass  of  the  smaller  circle),  and 
M^  =  TV  (a^  —  6^)  (the  mass  of  the  ring). 


Substituting  in  the  formula,  we  have 


Fig.  105 


whence,  by  solving  for  x^, 


7ra2  .  0  =  irh'^c  +  ir  {a^  -  h'^)  x^ 


It  is  unnecessary  to  find 
lies  on  OX. 


since,  by  symmetry,  the  center  of  gravity 


EXERCISES 


1.  Show  that  if  there  are  only  two  component  masses  M^  and  M^ 
in  formulas  (1)  of  the  theorem,  the  center  of  gravity  of  the  composite 
mass  lies  on  the  line  connecting  the  centers  of  gravity  of  the  com- 
ponent masses  at  such  a  point  as  to  divide  that  line  into  segments 
inversely  proportional  to  the  masses. 

2.  Prove  that  if  a  mass  M^  with  center  of  gravity  (x^^,  y^  has  cut 
out  of  it  a  mass  M^  with  center  of  gravity  (x^^  y^,  the  center  of  gravity 
of  the  remaining  mass  is 


M^x^ 


M,-M^ 


y 


M^  -  M^ 


r^  and  r^  are  tangent  externally.    Find 


3.  Two  circles  of  radii 
their  center  of  gravity. 

4.  Eind.  the  center  of  gravity  of  a  hemispherical  shell  bounded 
by  two  concentric  hemispheres  of  radii  r^  and  r^. 

5.  Place  r^  =  r J  +  Ar  in  Ex.  4,  let  Ar  approach  zero,  and  thus  find 
the  center  of  gravity  of  a  hemispherical  surface. 


CENTER  OF  GRAVITY  257 

6.  Find  the  center  of  gravity  of  a  hollow  right  circular  cone 
bounded  by  two  parallel  conical  surfaces  of  altitudes  h^  and  h^ 
respectively  and  with  their  bases  in  the  same  plane. 

7.  Place  7^2=  h^-\-  ^h  in  Ex.  6,  let  AA  approach  zero,  and  thus 
find  the  center  of  gravity  of  a  conical  surface. 

8.  Find  the  center  of  gravity  of  a  carpenter's  square  each  arm 
of  which  is  15  in.  on  its  outer  edge  and  2  in.  wide. 

9.  From  a  square  of  edge  8  in.  a  quadrant  of  a  circle  is  cut  out, 
the  center  of  the  quadrant  being  at  a  corner  of  the  square  and  the 
radius  of  the  quadrant  being  4  in.  Find  the  center  of  gravity  of  the 
figure  remaining. 

10.  Two  iron  balls  of  radius  4  in.  and  6  in.  respectively  are  con- 
nected by  an  iron  rod  of  length  1  in.  Assuming  that  the  rod  is  a 
cylinder  of  radius  1  in.,  find  the  center  of  gravity  of  the  system. 

11.  A  cubical  pedestal  of  side  4  ft.  is  surmounted  by  a  sphere  of 
radius  2  ft.  Find  the  center  of  gravity  of  the  system,  assuming 
that  the  sphere  rests  on  the  middle  point  of  the  top  of  the  pedestal. 

87.  Theorems.    The  following  theorems  involving  the  center 
of  gravity  may  often  be  used  to  ad- 
vantage  in  finding  pressures,  volumes 
of  solids    of   revolution,   or  areas  of 
surfaces  of  revolution. 

I.  The  total  pressure  on  a  plane  sur- 
face immersed  in  liquid  in  a  vertical 
position  is  equal  to  the  area  of  the  sur- 
face multiplied  hy  the  pressure  at  its 
center  of  gravity. 

Let  the  area  be  placed  as  in  Fig.  106,  p^^  j^g 

where  the  axis  of  x  is  in  the  surface 

of  the  liquid  and  where  the  axis  of  y  is  measured  downward. 
Then,  by  §  25, 

F=jwy{x^-x;)dy,  (1) 

which  may  be  written  as  a  double  integral  in  the  form 

p  =  I  /  wydydx z=w  \  i  ydxdy.  (2) 


258 


KEPEATED  INTEGRATION 


In  fact,  this  may  be  written  down  directly,  since  the  pressure  on 
a  small  rectangle  dxdy  is  its  area,  dxdy,  times  its  depth,  y,  times  w. 
Moreover,  from  §  85,  we  have 


Ay 


--  j  j  ydxdy. 


(3) 


y=a 


Fig.  107 


By  comparison  of  (2)  and  (3)  we  have 

P  =  wyA. 

But  wy  is  the  pressure  at  the  center  of  gravity,  and  the  theorem 
is  proved  for  areas  of  the  above  general  shape.    If  the  area  is 
not  of  this  shape,  it  may  be  divided  into  such  areas,  and  the 
theorem  may  be  proved  with  the  aid 
of  the  theorem  of  §  86. 

Ex.  1.    A  circular  bulkhead  which  closes    y=b 
the  outlet  of  a  reservoir  has  a  radius  3  ft., 
and  its  center  is  12  ft.  below  the  surface  of 
the  water.    Find  the  total  pressure  on  it. 

Here  ^  =  9  tt  and  the  depth  of  the  center 
of  gravity  is  12.   Therefore 

P  =  108  TTw  =  -2/-  '^  ^ons  =  10.6  tons. 

II.   77ie  volume  generated  by  revolving 
a  plane  area  about  an  axis  in  its  plane 

not  intersecting  the  area  is  equal  to  the  area  of  the  figure  multiplied 
by  the  circumference  of  the  circle  described  by  its  center  of  gravity. 

To  prove  this  take  an  area  as  in  Fig.  107.    Then,  by  §  26,  if 
V  is  the  volume  generated  by  the  revolution  about  OY, 

V=iT  [  (xl-x^dy, 

which  can  be  written  as  a  double  integral  in  the  form 

V—  2  7r  I      I     xdydx. 

By  §85,  Ax=  j     I     xdydx; 

and,  by  comparison  of  (4)  and  (5),  we  have 

V^^irxA, 
which  was  to  be  proved. 


(4) 


(5) 


THEOREMS  OF  PAPPUS  259 

Ex.  2.  Find  the  volume  of  the  ring  surface  formed  by  revolving  about 
an  axis  in  its  plane  a  circle  of  radius  a  whose  center  is  at  a  distance  c  from 
the  axis,  where  c>a. 

We  know  that  A  =  Tra^  and  that  the  center  of  gravity  of  the  circle  is 
at  the  center  of  the  circle  and  therefore  describes  a  circumference  of  length 
2  TTC.    Therefore  V=2  tt  Vc. 

III.  The  area  generated  hy  revolving  a  plane  curve  about  an 
axis  in  its  plane  not  intersecting  the  curve  is  equal  to  the  length 
of  the  curve  multiplied  hy  the  circumference  of  the  circle  described 
by  its  center  of  gravity. 

To  prove  this  we  need  a  formula  for  the  area  of  a  surface  of 
revolution  which  has  not  been  given.  It  may  be  shown  that  if 
S  is  this  area,  then  ^ 

S=2ir\xds.  (6) 

A  rigorous  proof  of  this  will  not  be  given  here.  However,  the 
student  may  make  the  formula  seem  plausible  by  noticing  that 
an  element  ds  of  the  curve  will  generate  on  the  surface  a  belt 
of  width  ds  and  length  2  irx.  The  product  of  length  by  breadth 
may  be  taken  as  the  area  of  the  belt. 

Moreover,  by  §  85,  we  have 

sx=  I  xds;  (7) 

and  comparing  the  two  equations  (6)  and  (7),  we  have 

S=27rsx, 
which  was  to  be  proved. 

Ex.  3.    Find  the  area  of  the  ring  surface  described  in  Ex.  2. 

We  know  that  s  =  2  7ra  and  that  the  center  of  gravity  of  a  circumfer- 
ence is  at  its  center  and  therefore  describes  a  circumference  of  length  2  ire. 
Therefore  g  =  4^  ^z^c. 

Theorems  II  and  III  are  known  as  the  theorems  of  Pappus. 

EXERCISES 

1.  Find  by  the  theorems  of  Pappus  the  volume  and  the  surface 
of  a  sphere. 

2.  Find  by  the  theorems  of  Pappus  the  volume  and  the  lateral 
surface  of  a  right  circular  cone. 


260  REPEATED  INTEGRATIOK 

3.  Find  by  the  theorems  of  Pappus  the  volume  generated  by 
revolving  a  parabolic  segment  about  its  altitude. 

4.  Find  by  the  theorems  of  Pappus  the  volume  generated 
by  revolving  a  parabolic  segment  about  its  base. 

5.  Find  by  the  theorems  of  Pappus  the  volume  generated  by 
revolving  a  parabolic  segment  about  the  tangent  at  its  vertex. 

6.  Find  the  volume  and  the  surface  generated  by  revolving  a 
square  of  side  a  about  an  axis  in  its  plane  perpendicular  to  one  of 
its  diagonals  and  at  a  distance  h  from  its  center. 

7.  Find  the  volume  and  the  area  generated  by  revolving  a  right 
triangle  with  legs  a  and  h  about  an  axis  in  its  plane  parallel  to  the 
leg  of  length  a  and  at  a  distance  c  from  the  vertex  of  the  right  angle 
on  the  opposite  side  from  the  hypotenuse. 

8.  A  circular  water  main  has  a  diameter  of  4  ft.  One  end  is 
closed  by  a  bulkhead,  and  the  other  is  connected  with  a  reservoir 
in  which  the  surface  of  the  water  is  18  ft.  above  the  center  of  the 
bulkhead.    Find  the  pressure  on  the  bulkhead. 

9.  Find  the  pressure  on  an  ellipse  of  semiaxes  a  and  h  completely 
submerged,  if  the  center  of  the  ellipse  is  c  units  below  the  surface  of 
the  liquid. 

10.  Find  the  pressure  on  a  semiellipse  of  semiaxes  a  and  h  {a  >  h) 
submerged  with  the  major  axis  in  the  surface  of  the  liquid  and  the 
minor  axis  vertical. 

11.  Find  the  pressure  on  a  parabolic  segment  submerged  with  the 
base  horizontal,  the  axis  vertical,  the  vertex  above  the  base,  and  the 
vertex  c  units  below  the  surface  of  the  liquid. 

12.  What  is  the  effect  on  the  pressure  of  a  submerged  vertical  area 
in  a  reservoir  if  the  level  of  the  water  in  the  reservoir  is  raised  by  c  feet  ? 

88.  Moment  of  inertia.  The  moment  of  inertia  of  a  particle 
about  an  axis  is  the  product  of  its  mass  and  the  square  of  its 
distance  from  the  axis.  The  moment  of  inertia  of  a  number 
of  particles  about  the  same  axis  is  the  sum  of  the  moments  of 
inertia  of  the  separate  particles  about  that  axis.  From  these 
definitions  we  may  derive  the  moment  of  inertia  of  a  thin  plate. 

Let  the  surface  of  the  plate  be  divided  into  elements  of 
area  dA.  Then  the  mass  of  each  element  is  pdA^  where  p  is  the 
product  of  the  thickness  of  the  plate  and  its  density.    Let  R  be 


MOMENT  OF  INERTIA 


261 


the  distance  of  any  point  in  the  element  from  the  axis  about 
which  we  wish  the  moment  of  inertia.  Then  the  moment  of 
inertia  of  element  is  approximately 

R'^p  dA, 

We  say  "  approximately  "  because  not  all  points  of  the  element 
are  exactly  a  distance  R  from  the  axis,  as  R  is  the  distance  of 
some  one  point  in  the  element.  However,  the  smaller  the  ele- 
ment the  more  nearly  can  it  be  regarded  as  concentrated  at  one 
point  and  the  limit  of  the  sum  of  all  the  elements,  as  their 
size  approaches  zero  and  their  number  increases  without  limit, 
is  the  moment  of  inertia  of  the  plate.  Hence,  if  /  represents  the 
moment  of  inertia  of  the  plate,  we  have 

/=  CR^pdA.  (1) 

If  in  (1)  we  let  /o  =  1,  the  resulting  equation  is 


=/^' 


dA, 


(2) 


where  /  is  called  the  moment  of  inertia  of  the  plane  area.  When 
dA  in  (1)  or  (2)  is  replaced  by  dxdi/  or  rdrdd,  the  double 
sign  of  integration  must  be  used. 

Ex.  1.  Find  the  moment  of  inertia  of  a  rectangle  of  dimensions  a  and  b 
about  the  side  of  length  b. 

Let  the  rectangle  be  placed  as  in  Fig.  108.  Let  it  be  divided  up  into 
elements  dA  =  dxdy.  Then  x  is  the  distance  of  some  point  in  an  element 
from   OY.    Hence,  in  (2),  we  have  y 

R  =  X  and  dA  =  dxdy.    Therefore 

/  =  iix^dxdy. 

We  first  sum  the  rectangles  in 
a  vertical  strip,  as  y  ranges  from  0 
to  b.   We  have 

I     x^dxdy  =  X'bdx. 

This  is  the  moment  of  inertia  of 
the  strip  MN,  and  might  have  been 
written  down  at  once,  since  all  points  on  the  left-hand  boundary  of  the 
strip  are  at  a  distance  x  from  OY  and  since  the  area  of  the  strip  is  bdx. 


262  REPEATED  INTEGRATION 

The  second  integration  gives  now 

1=  Cxndx  =  \a%. 

If,  instead  of  asking  for  the  moment  of  inertia  of  the  area,  we  had  asked 

for  that  of  a  plate  of  metal  of  thickness  t  and  density  D,  the  above  result 

would  be  multiplied  by  p  =  Dt.   But  in  that  case  the  total  mass  M  of  the 

plate  is  pah,  so  that  we  have 

^  ^  1=1  Ma\ 

Ex.  2.  Find   the  moment   of   inertia   of   the  quadrant  of   an  ellipse 

—  4-  ^  =  1  (a>h)  about  its  major  axis. 
a^      b^ 

If  we  take  any  element  of  area  as  dxdy,  we  find  the  distance  of  its 

lower  edge  from  the'  axis  about  which  we  wish  the  moment  of  inertia 

to  be  y  (Fig.  109).    Hence  R  =  y  and 


.  CCy'^dxdy. 


It  will  now  be  convenient  to  sum  first 
with  respect  to  x,  since  each  point  of  a 
horizontal  strip  is  at  the  same  distance  from 
OX.    We  therefore  write 


■.ffy^dydi 


Fig.  109 


Now,  indicating  by  x^  the  abscissa  of  a  point  on  the  ellipse  to  distinguish 
it  from  the  general  x  which  is  that  of  a  point  inside  the  ellipse,  we  have 
for  the  first  integration 

I    ^y'^dydx  =  y^x-^dy  =  -y^Vb^  —  y^dy. 

For  the  second  integration 

I  =  j  f\''Vb^-y^dy. 
b^  0 

To  integrate,  place  y  =  bsm  <f>.    Then 

I  =  ah^  f~^sm^<j>cos'^d>d<l>  = 
Jo 

If,  instead  of  the  area,  we  consider  a  thin  plate  of  mass  M,  the  above 

result  must  be  multiplied  by  p,  where  M  =  I  irabp ;  whence 

I=^Mb^. 

The  polar  moment  of  inertia  of  a  plane  area  is  defined  as  the 
moment  of  inertia  of  the  area  about  an  axis  perpendicular  to 
its  plane.     This  may  also  be  called  conveniently  the  moment 


Trah- 


MOMENT  OF  INERTIA  263 

of  inertia  with  respect  to  the  point  in  which  the  axis  cuts  the 
plane  of  the  area,  for  the  distance  of  an  element  from  the  axis 
is  simply  its  distance  from  that  point.  Thus  we  may  speak, 
for  example,  of  the  polar  moment  of  inertia  with  respect  to 
an  axis  through  the  origin  perpendicular  to  the  plane  of  an 
area,  or,  more  concisely,  of  the  polar  moment  with  respect  to 
the  origin. 

If  the  area  is  divided  into  elements  dxdy^  and  one  point  in 
the  element  has  the  coordinates  (x^  y)^  the  distance  of  that 
point  from  the  origin  is  \/x^-^y^.  That  is,  in  (2),  if  we  place 
dA  =  dxdy  and  R'^=x^+  y"^^  we  shall  have  the  formula  for  the 
polar  moment  of  inertia  with  respect  to  the  origin.  Denoting 
this  by  /jj,  we  have 

I.^fJCx'+fydxdy.  (3) 

This  integral  may  be  split  up  into  two  integrals,  giving 

I,=  ff^dxdi/+fffdydx,  (4) 

where  the  change  in  the  order  of  the  differentials  in  the  two 
integrals  indicates  the  order  in  which  the  integration  may  be 
most  conveniently  carried  out. 

The  first  integral  in  (4)  is  the  moment  of  inertia  about  OY 
and  may  be  denoted  by  I^ ;  the  second  integral  is  the  moment 
of  inertia  about  OX  and  may  be  denoted  by  I^.  Therefore 
formula  (4)  may  be  written  as 

so  that  the  problem  of  finding  the  moment  of  inertia  may  be 
reduced  to  the  solving  of  two  problems  of  the  type  of  the 
first  part  of  this  section. 

Ex.  3.  Find  the  polar  moment  of  inertia  of  an  ellipse  with  respect  to 
the  origin. 

In  Ex.  2  we  found  I^  for  a  quadrant  of  the  ellipse.    For  the  entire 

ellipse  it  is  four  times  as  great,  since  moments  of  inertia  are  added  by 

definition.    Hence 

4  =  1  7ra6«. 


264 


KEPEATED  INTEGRATION 


By  a  similar  calculation 
Therefore 


^y  =  i  'r«'^^- 


If  the  area  is  replaced  by  a  plate  of  mass  M,  this  result  gives 

If  polar  coordinates  are  used,  the  element  of  area  is  rdOdr, 
and  the  distance  of  a  point  in  an  element  from  the  origin  is  r. 
Hence,  in  (2),  dA  =  rdddr  and  E  =  r,    Therefore 


=//" 


dOdr. 


(6) 


In  practice  it  is  usually  convenient  to  integrate  first  with 
respect  to  r,  holding  6  constant.  This  is,  in  fact,  to  find  the 
polar  moment  of  inertia  of  a  sector  with  vertex  at  0. 

Ex.  4.  Find  the  polar  moment  of  inertia  of  a  circle  with  respect  to  a 
point  on  its  circumference. 

Let  the  circle  be  placed  as  in  Fig.  110.  Its  equation  is  then  (Ex.  1,  §  51) 
r  =  2acos^,  where  a  is  the  radius.  If  we  take  any  element  rdddr  and 
find  Iq  for  all  elements  which  lie  in  the 
same  sector  with  it,  we  have  to  add  the 
elements  r^dOdr,  with  r  ranging  from  0  to 
Tj,  where  r^  is  the  value  of  r  on  the  circle  ; 
and  therefore  r^  =  2a  cos  $.    We  have 

'r^dBdr  =  \  r^dO  =  4  a^  cos^OdO. 


We  have  finally  to  sum  these  quantities, 
with  $  ranging  from  —  ^  to  ^ We  have 

IT 

lQ=n4:  a*  co8*dde  =  f  ira^ 


Fig.  110 


If  M  is  the  mass  of  a  circular  plate,  this  result,  multiplied  by  p,  gives 

/o  =  i  Ma'. 

Ex.  5.  Find  the  polar  moment  of  inertia  of  a  circle  with  respect  to  its 
center. 

Here  it  will  be  convenient  first  to  find  the  polar  moment  of  inertia  of  a 
ring  (Fig.  111).  We  integrate  first  with  respect  to  6,  keeping  r  constant. 
We  have 


i: 


r^drde  =  2Trr^dr, 


MOMENT  OF  INERTIA 


265 


which  is  the  approximate  area  of  the  ring  2  vrdr  multiplied  by  the 
square  of  the  distance  of  its  inner  circumference  from  the  center.  We 
then  have,  by  the  second  integration, 


C^irr^dr 


l7ra'- 


Fig.  Ill 


If  Af  is  the  mass  of  a  circular  plate,  this 
result,  multiplied  by  p,  gives 

/o  =  h  Ma\ 

The  moment  of  inertia  of  a  solid 
of  revolution  about  the  axis  of  revo- 
lution is  the  sum  of  the  moments  of 
inertia  of  the  circular  sections  about 
the  same  axis ;  that  is,  of  the  polar 
moments  of  inertia  of  the  circular  sections  about  their  centers. 
If  the  axis  of  revolution  is  OF,  the  radius  of  any  circular 
section  perpendicular  to  Or  is  a;  and  its  thickness  is  dy.  Its 
mass  is  therefore  pirx^dy ;  and  therefore,  by  Ex.  5,  its  moment 
of  inertia  about  0  F  is  i  pirx^dy.  The  total  moment  of  inertia  of 
the  solid  is  therefore 

1=19-^  i  ^^dy. 

Ex.  6.    Find  the  moment  of  inertia  of  a  circular  cone  about  its  axis. 
Take  the  cone  as  in  Ex.  4,  §  85.    Then  we  have 
1        r  ^  a*  1 

But,  if  M  is  the  mass  of  the  cone,  we  have  M  =  \  pira^b. 


Therefore 


EXERCISES 

1.  Find  the  moment  of  inertia  of  a  rectangle  of  base  b  and  alti- 
tude a  about  a  line  through  its  center  and  parallel  to  its  base. 

2.  Find  the  moment  of  inertia  of  a  triangle  of  base  h  and  altitude 
a  about  a  line  through  its  vertex  and  parallel  to  its  base. 

3.  Find  the  moment  of  inertia  of  a  triangle  of  base  h  and  alti- 
tude a  about  its  base. 

4.  Find  the  moment  of  inertia  of  an  ellipse  about  its  minor  axis 
and  also  about  its  major  axis. 


266  KEPEATED  INTEGRATION 

6.  Find  the  moment  of  inertia  of  a  trapezoid  about  its  lower  base, 
taking  the  lower  base  as  h,  the  upper  base  as  a,  and  the  altitude  as  h. 

6.  Find  the  moment  of  inertia  about  its  base  of  a  parabolic  seg- 
ment of  base  h  and  altitude  a. 

7.  Find  the  polar  moment  of  inertia  of  a  rectangle  of  base  b  and 
altitude  a  about  its  center. 

8.  Find  the  polar  moment  of  inertia  about  its  center  of  a  circular 
ringj  the  outer  radius  being  r^  and  the  inner  radius  r^. 

9.  Find  the  polar  moment  of  inertia  of  a  right  triangle  of  sides 
a  and  h  about  the  vertex  of  the  right  angle. 

10.  Find  the  polar  moment  of  inertia  about  the  origin  of  the  area 
bounded  by  the  hyperbola  xy  =  Q>  and  the  straight  line  cc  +  y— 7  =  0. 

11.  Find  the  polar  moment  of  inertia  about  the  origin  of  the  area 
bounded  by  the  curves  y  =  x^  and  y  =  2  —  x^. 

12.  Find  the  polar  moment  of  inertia  about  the  origin  of  the  area 
of  one  loop  of  the  lemniscate  r^  =  2  a^  cos  2  6. 

13.  Find  the  moment  of  inertia  of  a  right  circular  cylinder  of 
height  hy  radius  r,  and  mass  M,  about  its  axis. 

14.  Find  the  moment  of  inertia  about  its  axis  of  a  hollow  right  cir- 
cular cylinder  of  mass  M,  its  inner  radius  being  r^,  its  outer  radius  r^, 
and  its  height  h. 

15.  Find  the  moment  of  inertia  of  a  solid  sphere  about  a  diameter. 

16.  A  ring  is  cut  fram  a  spherical  shell  whose  inner  and  outer  radii 
are  respectively  5  ft.  and  6  ft.,  by  two  parallel  planes  on  the  same 
side  of  the  center  and  distant  1  ft.  and  3  ft.  respectively  from  the 
center.    Find  the  moment  of  inertia  of  this  ring  about  its  axis. 

17.  The  radius  of  the  upper  base  and  the  radius  of  the  lower  base 
of  the  frustum  of  a  right  circular  cone  are  respectively  r^  and  r^,  and 
its  mass  is  M.    Find  its  moment  of  inertia  about  its  axis. 

89.  Moments  of  inertia  about  parallel  axes.  The  finding  of  a 
moment  of  inertia  is  often  simplified  by  use  of  the  following 
theorem : 

The  moment  of  inertia  of  a  body  about  an  axis  is  equal  to  its 
moment  of  inertia  about  a  parallel  axis  through  its  center  of  gravity 
plus  the  product  of  the  mass  of  the  body  by  the  square  of  the 
distance  between  the  axes. 


MOMENT  OF  INERTIA 


267 


We  shall  prove  this  theorem  only  for  a  plane  area,  in  the 
two  cases  in  which  the  axes  lie  in  the  plane  of  the  figure  or 
are  perpendicular  to  that  plane.  We  shall  also  consider  the 
mass  of  the  area  as  equal  to  the 
area,  as  in  §  88. 

Case  I,    When  the  axes  lie  in  the 
plane  of  the  figure. 

Let  the  area  be  placed  as  in 
Fig.  112,  where  the  center  of  grav- 
ity (x,  ^)  is  taken  as  the  origin 
(0,  0)  and  where  the  axis  of  ^  is 
taken  parallel  to  the  axis  LK, 
about  which  we  wish  to  find  the 
moment  of  inertia.  Let  x  be  the  distance  of  an  element  dxdi/ 
from  OF,   and  x^   its   distance    from  LK.    Then,   if  /    is   the 


Fig.  112 


moment  of  inertia  about  OF,  and  I^  the  moment  of  inertia  about 
LK,  we  have  ^^  ^^ 


:  /  i  x'^dxdy. 


Moreover,  if  a  is  the  distance  between  0  F  and  LK,  we  have 

x^=x-{-a\  (2) 

so  that,  by  substituting  from  (2)  in  the  second  equation  of  (1), 
we  have  ^^  ^^ 

Ii=  I  f  x^dxdi/  -\-2a  I  I  xdxdy  -{•  a^  I  I  dxdy,  (3) 


Now,  by  §  84,  CCdxdy  =  ^  ;  by  §  85,  CC xdxdy  =  Ax=^, 


smce 


by  hypothesis  2:  =  0  ;  and,  by  (1),  the  first  integral  on  the  right 
hand  of  (3)  is  I^.    Therefore  (3)  can  be  written 


Z  =  J  +  aM, 


(4) 


which  proves  the  theorem  for  this  case. 

Case  IL  When  the  axes  are  perpendicular  to  the  plane  of 
the  figure. 

We  have  to  do  now  with  polar  moments  of  inertia.  Let  the 
area  be  placed  as  in  Fig.  113,  where  the  center  of  gravity  is 


268 


REPEATED  INTEGRATION 


taken  as  the  origin,  and  P  is  any  point  about  which  we  wish 
the  polar  moment  of  inertia.  Let  Ig  be  the  polar  moment  of 
inertia  about  0,  and  I^  the  polar  moment  of  inertia  about  P. 
Draw  through  P  axes  PX'  and 
P  Y'  parallel  to  the  axes  of  coor- 
dinates OX  and  0  Y,  Let  I^  and 
/y  be  the  moments  of  inertia 
about  OX  and  OY  respectively, 
and  let  I^,  and  I^,  be  the  moments 
of  inertia  about  PX'  and  PY', 
Then,  by  (5),  §  88, 


Fig.  113 


Moreover,  if  (a,   5)  are  the  coordinates  of  P,  we  have,  by 
Case  I,  r_r_L^2j  /,  =  /-f  h'^A, 


i,=  r  +  d^A, 


(6) 


Therefore,  from  (5),  we  have 

I,  =  I,+  (ia'+h')A,  (7) 

which  proves  the  theorem  for  this  case  also. 

The  student  may  easily  prove  that  the  theorem  is  true  also 
for  the  moment  of  inertia  of  any  solid  of  revolution  about  an 
axis  parallel  to  the  axis  of  revolution  of  the  solid. 

Ex.  Find  the  polar  moment  of  inertia  of  a  circle  with  respect  to  a  point 
on  the  circumference. 

The  center  of  gravity  of  a  circle  is  at  its  center,  and  the  distance  of  any 
point  on  its  circumference  from  its  center  is  a.  By  Ex.  5,  §  88,  the  polar 
moment  of  a  circle  about  its  center  is  ^tto*.  Therefore,  by  the  above 
theorem,  ^        ,      a  ,     o  ^     o^       ^      a 

This  result  agrees  with  Ex.  4,  §  88,  where  the  required  moment  of  inertia 
was  found  directly. 

EXERCISES 

1.  Find  the  moment  of  inertia  of  a  circle  about  a  tangent. 

2.  Find  the  polar  moment  of  inertia  about  an  outer  corner  of 
a  picture  frame  bounded  by  two  rectangles,  the  outer  one  being  of 
dimensions  8  ft.  by  12  ft.,  and  the  inner  one  of  dimensions  5  ft.  by  9  ft. 


SPACB  COORDINATES  269 

3.  Find  the  moment  of  inertia  about  one  of  its  outer  edges  of  a 
carpenter's  square  of  which  the  outer  edges  are  15  in.  and  the  inner 
edges  13  in. 

4.  Find  the  polar  moment  of  inertia  about  the  outer  corner  of 
the  carpenter's  square  in  Ex.  3. 

5.  From  a  square  of  side  20  a  circular  hole  of  radius  5  is  cut, 
the  center  of  the  circle  being  at  the  center  of  the  square.  Find  the 
moment  of  inertia  of  the  resulting  figure  about  a  side  of  the  square. 

6.  Find  the  polar  moment  of  inertia  about  a  corner  of  the  square 
of  the  figure  in  Ex.  5. 

7.  Find  the  moment  of  inertia  of  a  hollow  cylindrical  column 
of  outer  radius  r^  and  inner  radius  r^  about  an  element  of  the  inner 
cylinder, 

8.  Find  the  moment  of  inertia  of  the  hollow  column  of  Ex.  7 
about  an  element  of  the  outer  cylinder. 

9.  Find  the  moment  of  inertia  of  a  circular  ring  of  inner  radius  r^ 
and  outer  radius  r^  about  a  tangent  to  the  outer  circle. 

10.  A  circle  of  radius  a  has  cut  from  it  a  circle  of  radius  -  tangent 

z 

to  the  larger  circle.    Find  the  moment  of  inertia  of  the  remaining 
figure  about  the  line  through  the  centers  of  the  two  circles. 

11.  Find  the  moment  of  inertia  of  the  figure  in  Ex.  10  about  a 
line  through  the  center  of  the  larger  circle  perpendicular  to  the  line 
of  centers  of  the  two  circles  and  in  the  plane  of  the  circles. 

90.  Space  coordinates.  In  the  preceding  pages  we  have  be- 
come familiar  with  two  methods  of  fixing  the  position  of  a 
point  in  a  plane ;  namely,  by  Cartesian  coordinates  (x^  y)^  and 
by  polar  coordinates  (r,  ^).  If,  now,  any  plane  has  been  thus 
supplied  with  a  coordinate  system,  and,  starting  from  a  point 
in  that  plane,  we  measure  another  distance,  called  «,  at  right 
angles  to  the  plane,  we  can  reach  any  point  in  space.  The  quan- 
tity z  will  be  considered  positive  if  measured  in  one  direction, 
and  negative  if  measured  in  the  other.  We  have,  accordingly, 
two  systems  of  space  coordinates. 

1.  Cartesian  coordinates.  We  take  any  plane,  as  XOY^  in  which 
are  already  drawn  a  pair  of  coordinate  axes,  OX  and  OF,  at 
right  angles  with  each  other.    Perpendicular  to  this  plane  at 


270 


KEPEATED  INTEGRATION 


the  origin  we  draw  a  third  axis  OZ  (Fig.  114).  If  P  is  any 
point  of  space,  we  draw  PM  parallel  to  OZ,  meeting  the  plane 
XO  Y  at  M^  and  from  M  draw  a  line  par- 
allel to  OY^  meeting  OX  at  L,  Then  for 
the  point  P  (x,  y^  2),  OL  —  x^  LM=  y, 
and  MP  =  z.  It  is  to  be  noticed  that 
the  three  axes  determine  three  planes, 
XOY,  YOZ,  and  ZOX,  called  the  coor- 
dinate planes,  and  that  we  may  just  as 
readily  draw  the  line  from  P  perpendic- 
ular to  either  the  plane  YOZ  or  ZOX  and 
then  complete  the  construction  as  above. 

These  possibilities  are  shown  in  Fig.  115,  where  it  is  seen  that 
x=  0L  =  NM=  SB  =  TP,  with  similar  sets  of  values  for  ?/  and  z. 

2.  Cylindrical  coordinates.  Let  XOY 
be  any  plane  in  which  a  fixed  point  O 
is  the  origin  of  a  system  of  polar  coor- 
dinates, and  OX  is  the  initial  line  of  that 
system  (Fig.  116).  Let  OZ  be  an  axis 
perpendicular  to  the  plane  XOY  at  0. 
If  P  is  any  point  in  space,  we  draw 
from  P  a  straight  line  parallel  to  OZ 
until  it  meets  the  plane  XOY  at  31, 
Then,  if  the  polar  coordinates  of  M  in 

the  plane  XO  Y  are  r  =  OM,  6  =  XOM^  and  we  denote  the  dis- 
tance MP  by  2,  the  cylindrical  coordinates  of  P  are  (r,  6,  2;). 

It  is  evident  that  the  axes  OX  and 
OZ  determine  a  fixed  plane,  and  that 
the  angle  6  is  the  plane  angle  of  the 
dihedral  angle  between  that  fixed  plane 
and  the  plane  through  OZ  and  the 
point  P,  If  SP  is  drawn  in  the  latter 
plane  perpendicular  to  OZ,  it  is  evident 
that  OM=SP  =  r  and  OS  =  MP  =  z. 
The  coordinate  r,  therefore,  measures 
the  distance  of  the  point  P  from  the  axis  OZ,  and  the  coordinate 
z  measures  the  distance  of  P  from  the  plane  XOY, 


Fig.  116 


SURFACES 


271 


If  the  line  OX  of  the  cylindrical  coordinates  is  the  same  as 
the  axis  OX  of  the  Cartesian  coordinates,  and  the  axis  OZ  is  the 
same  in  both  systems,  it  is  evident,  from  §  51,  that 

x  —  r  cos  ^,         1/  =  r  sin  ^,         2  =  z.  (1) 

These  are  formulas  by  which  we  may  pass  from  one  system 
to  the  other.    It  is  convenient  to  notice  especially  that 

r^=2^+/.  (2) 

91.  Certain  surfaces.  A  single  equation  between  the  coordi- 
nates of  a  point  in  space  represents  a  surface.  We  shall  give 
examples  of  the  equations  of  certain  surfaces  which  are  impor- 
tant in  applications.  In  this  connection  it  should  be  noticed 
that  when  we  speak  of  the  equation  of  a  sphere  we  mean  the 
equation  of  a  spherical  surface,  and  when  we  speak  of  the 
volume  of  a  sphere  we  mean  the  volume  of  the  solid  bounded 
by  a  spherical  surface.  The  word  sphere^  then,  indicates  a  sur- 
face or  a  solid,  according  to  the  context.  Similarly,  the  word 
cone  is  used  to  denote  either  a  conical  surface  indefinite  in 
extent  or  a  solid  bounded  by  a  conical  surface  and  a  plane 
base.  It  is  in  the  former  sense  that  we  speak  of  the  equation 
of  a  cone,  and  in  the  latter  sense 
that  we  speak  of  the  volume  of 
a  cone.  In  the  same  way  the 
word  cylinder  may  denote  either 
a  cylindrical  surface  or  a  solid 
bounded  by  a  cylindrical  surface 
and  two  plane  bases.  This  double 
use  of  these  words  makes  no  con- 
fusion in  practice,  as  the  context 
always  indicates  the  proper  mean- 
ing in  any  particular  case. 

1.  Sphere  with  center  at  origin.  Con- 
sider any  sphere  (Fig.  117)  with  its  center 
at  the  origin  of  coordinates  and  its  radius 
equal  to  a.  Let  P  be  any  point  on  the  surface  of  the  sphere.  Pass  a  plane 
through  P  and  OZ,  draw  PS  perpendicular  to  OZ,  and  connect  0  and  P. 
Then,  using  cylindrical  coordinates,  in  the  right  triangle  OPS,  OS=z,  SP=r, 


Fig.  117 


and  OP  =  a.   Therefore 


r^+  z^=  a\ 


(1) 


272 


EEPEATED  INTEGRATION 


Fig. 118 


This  equation  is  satisfied  by  the  cylindrical  coordinates  of  any  point  on 
the  surface  of  the  sphere  and  by  those  of  no  other  point.    It  is  therefore 
the  equation  of  the  sphere  in  cylindrical 
coordinates.  Z 

By  means  of  (2),  §  90,  equation  (1) 

becomes 

a:2  +  2/2  +  22  =  a\  (2) 

which  is  the  equation  of  the  sphere  in 
Cartesian  coordinates. 

2.  Sphere  tangent  at  origin  to  a  coor- 
dinate plane.  Consider  a  sphere  tangent 
to  the  plane  XOY  at  O  (Fig.  118).  Let 
P  be  any  point  on  the  surface  of  the 
sphere.  Let  A  be  the  point  in  which  the 
axis  OZ  again  meets  the  sphere.  Pass  a 
plane  through  P  and  OZ  ;  connect  A  and 
P,  0  and  P ;  and  draw  PS  perpendicular 
to  OZ.  Then,  using  cylindrical  coordi- 
nates, OS  =  z,  SP  =  r,  and  OA  =2a, 
where   a  is  the   radius   of   the  sphere. 

Now  OAP  is  a  right  triangle,  since  it  is  inscribed  in  a  semicircle, 
and  PS  is  the  perpendicular  from  the  vertex  of  the  right  angle  to  the 
hypotenuse.  Therefore,  by  elementary  plane  geometry, 

'SP^=OS-SA  =OS{OA  -OS). 

Substituting  the  proper  values,  we  have 

r^=2az-z'^,  (3) 

which  is  the  equation  of  the  sphere  in  cylindrical 
coordinates. 

By  (2),  §  90,  equation  (3)  becomes 

x^-\-  f-\-  z^-2az  =  0,  (4) 

which  is  the  equation  of  the   sphere  in    Cartesian 
coordinates. 

3.  Right  circular  cone.    Consider  any  right  circular 
cone  with  its  vertex  at  the  origin  and  its  axis  along 
OZ  (Fig.  119).   Let  a  be  the  angle  which  each  element 
of  the  cone  makes  with  OZ.    Take  /-*  any  point  on  the  surface  of  the  cone, 
pass  a  plane  through  P  and  OZ,  and  draw  PS  perpendicular  to  OZ.   Then 

SP 


Fig. 119 


SP  =  r  and  05  =  z.    But  —-  =  tan  SOI 
OS 


tana.    Therefore  we  have 
r  =  z  tan  a  (5) 

as  the  equation  of  the  cone  in  cylindrical  coordinates. 


SURFACES 


273 


By  2,  §  90,  equation  (5)  becomes 

x^-{-  y^-z^tan^a  =  0,  (6) 

as  the  equation  of  the  cone  in  Cartesian  coordinates. 

As  explained  above,  we  have  here  used  the  word  cone  in  the  sense  of  a 
conical  surface.   If  the  cone  is  a  solid  with  its  altitude  h  and  the  radius  of 

its  base  a,  then  tan  a  =  -  •  In  this  case  equation 
h 

(5)  or  (6)  is  that  of  the  curved  surface  of  the 
cone  only. 

4.  Surface  of  revolution.  Consider  any  sur- 
face of  revolution  with  OZ  the  axis  of  revolution 
(Fig.  120).  Take  P  any  point  on  the  surface 
and  pass  a  plane  through  P  and  OZ.  In  the 
plane  POZ  draw  OR  perpendicular  to  OZ  and, 
from  P,  a  straight  line  perpendicular  to  OZ 
meeting  OZ  in  S.  If  we  regard  OR  and  OZ  as 
a  pair  of  rectangular  axes  for  the  plane  POZ, 
the  equation  of  the  curve  CD  in  which  the 
plane  POZ  cuts  the  surface  is  Fig.  120 

^  =f(r),  (7) 

exactly  as  y  =f(x)  is  the  equation  of  a  curve  in  §  12. 

But  CD  is  the  same  curve  in  all  sections  of  the  surface  through  OZ. 
Therefore  equation  (7)  is  true  for  all  points  P  and  is  the  equation  of  the  sur- 
face in  cylindrical  coordinates.  When  the  plane  POZ  coincides  with  the 
plane  X  OZ,  r  is  equal  to  x,  and  equation  (7)  becomes,  y 

for  that  section,         ^  _  f(x).  (S) 

Hence  we  have  the  following  theorem  : 

The  equation  of  a  surface  of  revolution  formed  hy 
revolving  about  OZ  any  curve  in  the  plane  XOZ  may 
be  found  in  cylindrical  coordinates  by  writing  r  for  x 
in  the  equation  of  the  curve. 

The  equation  of  the  surface  in  Cartesian  coor- 
dinates may  then  be  found  by  placing  r  =  Vx^  -t-  y'^. 
For  example,  the  equation  of  the  surface  formed 
by  revolving  the  parabola  z^=  ix  about  OZ  as 
an  axis  is  z^=  ir  in  cylindrical  coordinates,  or 
z*  =  IQ  (x^  ■{■  y^)  in  Cartesian  coordinates. 

5.  Cylinder.  Consider  first  a  right  circular  cylinder  with  its  axis  along 
OZ  (Fig.  121).  From  any  point  P  of  the  surface  of  the  cylinder  draw  PS 
perpendicular  to  OZ.  Then  SP  is  always  equal  to  a,  the  radius  of  the 
cylinder.    Therefore,  for  all  points  on  the  surface, 

r  =  a,  (9) 


274 


KEPEATED  INTEGRATION 


which  is  the  equation  of  the  cylinder  in  cylindrical  coordinates.    Reduced 
to  Cartesian  coordinates  equation  (9)  becomes 

X^+y2=a\  (10) 

the  equation  of  the  cylinder  in  Cartesian  coordinates. 

More  generally,  any  equation  in  x  and  y  only,  or  in  r  and  6  only,  repre- 
sents a  cylinder.  In  fact,  either  of  these  equations,  if  interpreted  in  the 
plane  XOY,  represents  a  curve;  but  if  a  line  is  drawn  from  any  point  in 
this  curve  perpendicular  to  the  plane  XOY,  and  P  is  any  point  on  this  line, 
the  coordinates  of  P  also  satisfy  the  equation,  since  z  is  not  involved  in 
the  equation.  As  examples,  the  equation  y'^=  ^x  represents  a  parabolic 
cylinder,  and  the  equation  r  =  a  sin  3  ^  represents  a  cylinder  whose  base  is 
a  rose  of  three  leaves  (Fig.  65,  p.  144). 

6.  Ellipsoid.  Consider  the  surface  defined  by  the  equation 


+  P  +  ? 


=  1. 


ai) 


If  we  place  2  =  0,  we  get  the  points  on  the  surface  which  lie  in  the  XOY 
plane.   These  points  satisfy  the 
equation      o      ,,2 

and  therefore  form  an  ellipse. 

Similarly,  the  points  in  the 
ZOX  plane  lie  on  the  ellipse 

^^  +  1  =  1'  (1^) 

and  those  in  the  YOZ  plane  lie 
on  the  ellipse 


'/>2  ^  £.2 


1. 


(14) 


.--^ — J- — .^^-^ ,__ 

\^_^  "^ — ' 1^         ^^^ 

— -^-^^  \y    I      '_— -^' — 


Fig.  122 


The    construction    of    these 
ellipses  gives  a  general  idea  of 
the  shape  of  the  surface  (Fig.  122).    To  make  this  more  precise, 
place  z  —  z^  in  (11),  where  z^  is  a  fixed  value.    We  have 


which  can  be  written 


a'       IP- 


l-^> 


«^(^-S)  ^^(i-S) 


let  us 


(15) 


(16) 


which  is  satisfied  by  all  points  which  lie  in  the  plane  at  the  distance 
from  the  A' OF  plane. 


SURFACES  275 

As  long  as  z^  <  c^,  equation  (16)  represents  an  ellipse  with  semiaxes 


^■2 


a  v'l  —  ^  and  ^^  \'l  —  -^  •    ^J  taking  a  sufficient  number  of  these  sections 

we  may  construct  the  ellipsoid  with  as  much  exactness  as  desired. 

If  z^  —  c^  in  (16),  the  axes  of  the  ellipse  reduce  to  zero,  and  we  have  a 
point.    If  z^  >  c^  the  axes  are  imaginary,  and  there  is  no  section. 

7.  Elliptic  paraboloid.    Consider  the  surface 


?  =  ^  +  ^, 
c      a^      ¥ 


where  we  shall  assume,  for  definiteness,  that  c  is  positive. 


If  we  place  2  =  0,  we  get 


a''      V'        ' 


(17) 


(18) 


which  is  satisfied  in  real  quantities  only  by  a:  =  0  and  y  =  0.   Therefore  the 
XOY  plane  simply  touches  the  surface  at 
the  origin. 

If  we  place  z  =  c,  we  get  the  ellipse 


a^^  b'~^' 


(19) 


which  lies  in  the  plane  c  units  distant  from 
the  XOF  plane. 

If  we  place  y  =  0,  we  get  the  parabola 


^    2 

z  =  —  x^; 

a^ 


(20) 


and  if  we  place  a:  =  0,  we  get  the  parabola 
c 


(21) 


Fig.  123 


The  sections  (19),  (20),  and  (21)  determine  the  general  outline  of 
surface.    For  more  detail  we  place  z  =  z^  and  find  the  ellipse 

x^  V- 

^'^^"^'  (22) 

c  c 

so  that  all  sections  parallel  to  the  XOY  plane  and  above  it  are  ellipses 
(Fig.  123). 

8.  Elliptic  cone.    Consider  the  surface 


-  +  ^  -  -  =  0. 


(23) 


.  Proceeding  as  in  7,  we  find  that  the  section  r  =  0  is  simply  the  origin 
and  that  the  section  z  =  c  is  the  ellipse 


62 


1. 


(24) 


276 


REPEATED  INTEGRATION 


If  we  place  a;  =  0,  we  get  the  two  straight  lines 

h 

and  if  we  place  y  =  0,  we  get  the  two  straight  lines 


X  =  ±  -z. 
c 


(25) 


(26) 


The  sections  we  have  found  suggest  a  cone  with  an  elliptic  base.  To 
prove  that  the  surface  really  is  a  cone,  we  change  equation  (23)  to  cylin- 
drical coordinates,  obtaining 


fco^      sin^^' 


\  2       ^^ 


(27) 


Now  if  6  is  held  constant  in  (27),  the  coefficient  of  r^  is  constant,  and  the 


equation  may  be  written 


±  kz, 


(28) 


which  is  the  equation  of  two  straight  lines  in  the  plane  through  OZ 
Z  Z 


Fig.  124 


Fig.  125 


determined  by  ^  =  const.    Hence  any  plane  through  OZ  cuts  the  surface  in 
two  straight  lines,  and  the  surface  is  a  cone  (Fig.  124). 
9.  Plane.    Consider  the  surface 

Ax  +  By-{-  Cz  ■\-  D  =  0.  (29) 

The  section  2  =  0  is  the  straight  line  KH  (Fig.  125)  with  tlie  equation 
Ax-\-  By\-  D  =  0,  (30) 

the  section  ^  =  0  is  the  straight  line  LH  with  the  equation 

Ax  +  Cz  +  D  =  0,  (31) 

and  the  section  x  =  0  is  the  straight  line  LK  with  the  equation 

B,j+Cz  +  D^Q.  (32) 


VOLUME  277 

The  two  lines  (31)  and  (32)  intersect  OZ  in  the  point  L  (o,  0,  -  -), 

unless  C  =  0.   Assuming  for  the  present  that  C  is  not  zero,  we  change 
equation  (30)  to  cylindrical  coordinates,  obtaining 

(.4  cos  ^  +  5  sin  B)r-\-  Cz^  D  =  0.  (33) 

This  is  the  equation  of  a  straight  line  LN  in  the  plane  Q  —  const.  It 
passes  through  the  point  Z,  which  has  the  cylindrical  coordinates  r  =  0, 

2  =  —  — ;  and  it  meets  the  line  KHy  since  when  2  =  0,  equation  (33)  is  the 

same  as  equation  (30).    Hence  the  surface  is  covered  by  straight  lines  which 

pass  through  L  and  meet  KH.    The  locus  of  such  lines  is  clearly  a  plane. 

We  have  assumed  that  C  in  (29)  is  not  zero.   If  C  =  0,  equation  (29)  is 

^x  +  £?/  +  i)  =  0.  (34) 

The  point  L  does  not  exist,  since!  the  lines  corresponding  to  HL  and  KL 
are  now  parallel.  But,  by  5,  equation  (34)  represents  a  plane  parallel  to 
OZ  intersecting  XOY  va.  the  line  whose  equation  is  (34).  Therefore  we 
have  the  following  theorem  : 

Any  equation  of  the  Jirst  degree  represents  a  plane. 

92.  Volume.  Starting  from  any  point  (x^  y^  z)  in  space,  we 
may  draw  lines  of  length  dx,  dy^  and  dz  in  directions  parallel 
to  OXj  OY,  and  OZ  respectively,  and  on  these  lines  as  edges 
construct  a  rectangular  parallelepiped.  The  volume  of  this  fig- 
ure we  call  the  element  of  volume  dV  and  have 

dV=dxdydz.  (1) 

For  cylindrical  coordinates  we  construct  an  element  of  volume 
whose  base  is  rd6dr(^  84),  the  element  of  plane  area  in  polar 
coordinates,  and  whose  altitude  is  dz.  This  figure  has  for  its 
volume  d  V  the  product  of  its  base  by  its  altitude,  and  we  have 

dV=rdedrdz.  (2) 

The  two  elements  of  volume  dV  given  in  (1)  and  (2)  are 
not  equal  to  each  other,  since  they  refer  to  differently  shaped 
figures.  Each  is  to  be  used  in  its  appropriate  place.  To  find 
the  volume  of  any  solid  we  divide  it  into  elements  of  one  of 
these  types. 

To  do  this  in  Cartesian  coordinates,  note  that  the  a:-coordinate 
of  any  point  will  determine  a  plane  parallel  to  the  plane  YOZ 


278 


EEPEATED  INTEGRATION 


Fig.  126 


and  X  units  from  it,  and  that  similar  planes  correspond  to  the 
values  of  y  and  z.  We  may,  accordingly,  divide  any  required 
volume  into  elements  of  volume  as  follows: 

Pass  planes  through  the  volume  parallel  to  YOZ  and  dx 
units  apart.  The  result  is  to  divide  the  required  volume  into 
slices  of  thickness  dx^  one  of  which 
is  shown  in  Fig.  126.  Secondly, 
pass  planes  through  the  volume 
parallel  to  XOZ  and  dy  units  apart, 
with  the  result  that  each  slice  is 
divided  into  columns  of  cross  sec- 
tion dxdy.  One  such  column  is 
shown  in  Fig.  126. 

Finally,  pass  planes  through  the 
required  volume  parallel  to  XOY 
and  dz  units  apart,  with  the  result 
that  each  column  is  divided  into 
rectangular  parallelepipeds  of  dimensions  dx^  dy^  and  dz.  One 
of  these  is  shown  in  Fig.  126. 

It  is  to  be  noted  that  the  order  followed  in  the  above 
explanation  is  not  fixed  and  that,  in  fact,  the  choice  of  be- 
ginning with  either  x  oy  y  oy  z^  and  the  subsequent  order, 
depend  upon  the  particular  volume 
considered. 

A  similar  construction  may  be 
made  for  cylindrical  coordinates. 
In  this  case  the  coordinate  6 
determines  a  plane  through  OZ. 
We  accordingly  divide  the  volume 
by  means  of  planes  through  OZ 
making  the  angle  dQ  with  each 
other.  The  result  is  a  set  of  slices 
one  of  which  is  shown  in  Fig.  127. 

The  coordinate  r  determines  a  cylinder  with  OZ  as  its  axis. 
We  accordingly  divide  each  slice  into  columns  by  means  of 
cylinders  with  radii  differing  by  dr.  One  such  column  is  shown 
in  Fig.  127. 


Fig.  127 


VOLUME  279 

Finally,  these  columns  are  divided  into  elements  of  volume 
by  planes  parallel  to  XOY  at  a  distance  dz  apart.  One  such 
element  is  shown  in  Fig.  127. 

When  the  volume  has  been  divided  in  either  of  these  ways, 
it  is  evident  that  some  of  the  elements  will  extend  outside  the 
boundary  surfaces  of  the  solid.  The  sum  of  all  the  elements 
that  are  either  completely  or  partially  in  the  volume  will  be 
approximately  the  volume  of  the  solid,  and  this  approximation 
becomes  better  as  the  size  of  each  element  becomes  smaller. 
In  fact,  the  volume  is  the  limit  of  the  sum  of  the  elements. 
The  determination  of  this  limit  involves  in  principle  three  in- 
tegrations, and  we  write 

F=  CCCdxdydz  (3) 

V=  CCCrdOdrdz,  (4) 


or 


In  carrying  out  the  integrations  we  may,  in  some  cases,  find 
it  convenient  first  to  hold  z  and  dz  constant.  We  shall  then 
be  taking  the  limit  of  the  sum  of  the  elements  which  lie  in  a 
plane  parallel  to  the  XOY  plane.  We  may  indicate  this  by 
writing  (3)  or  (4)  in  the  form 

V=  fdz  CCdxdy      or     F=  C dz  ffrdddr.  (5) 

But,  by  §  84,  I  /  dxdy  =  A    and    j  I  rd6dr  =  A,  where  A  is 

the  area  of  the  plane  section  at  a  distance  z  from  XOY.    Hence 
(5)  is  ^ 

V=      Adz,  (6) 

in  agreement  with  §  26.  ^ 

Hence,  whenever  it  is  possible  to  find  A  by  elementary  means 
without  integration,  the  use  of  (6)  is  preferable.  This  is  illus- 
trated in  Ex.  1. 

In  some  cases,  however,  this  method  of  evaluation  is  not 
convenient,  and  it  is  necessary  to  carry  out  three  integrations. 
This  is  illustrated  in  Ex.  2. 


280 


REPEATED  INTEGRATION 


-p2  2/2  2" 

Ex.  1.  Find  the  volume  of  the  ellipsoid  -;,  +  ^  •¥  —  —  l. 

a^      b^      c^ 

By  6,  §  91,  the  section  made  by  a  plane  parallel  to  XOY  is  an  ellipse 

V"     ^                \        ^2 
1 and  i  a/1 ^  •    Therefore,  by  Ex.  1,  §  77,  its  area 

is  TTob  |1 ^1-    Hence  we  use  formula  (6)  and  have 

V  =  irab  j     (l -\  dz  =  -  irabc. 

Ex.  2.  Find  the  volume  bounded  above  by  the  sphere  x^i-  y^+  z^=  5  and 
below  by  the  paraboloid  x^+  y^  =  4iZ  (Fig.  128). 

As  these  are  surfaces  of  revolution,  this  example  may  be  solved  by 
the  method  of  Ex.  1,  but  in  so  doing  we  need  two  integrations  —  one 
for  the  sphere  and  the  other  for  the  paraboloid.  We  shall  solve  the 
example,  however,  by  the  other  method  in  order  to  illustrate  that  method. 

We  first  reduce  our  equations  to  cylin- 
drical coordinates,  obtaining  respectively 


r2  +  22 


and 


=  iz. 


(1) 
(2) 


The  surfaces  intersect  when  r  has  the 

same  value  in  both  equations;  that  is, 

when  1  ,  A         '  /ox 

2^  +  4  2  =  o,  (3) 

which  gives  z  =1ot  z  =  —  6.    The  latter 

value  is  impossible  ;  but  when  z  =  1,  we 

have  r  =  2  in  both  equations.   Therefore 

the  surfaces  intersect  in  a  circle  of  radius  2  in  the  plane  z 

lies  directly  above  the  circle  r  =  2  in  the  XOY  plane. 

We  now  imagine  the  element  rdOdrdz  inside  the  surface  and,  holding 
r,  B,  dO,  dr  constant,  we  take  the  sum  of  all  the  elements  obtained  by  varying 
z  inside  the  volume.    These  elements  obviously  extend  from  2  =  Zj  in  the 

lower  boundary  to  z  =  z^  in  the  upper  boundary.    From  (2),  z^  =  —  and, 


from  (1),  ^2  =  V  5  — 


•dSdrf^ 


The  first  integration  is  therefore 


Uddr 


We  must  now  allow  6  and  r  so  to  vary  as  to  cover  the  entire  circle  r  =  2 
above  which  the  required  volume  stands. 

If  we  hold  6  constant,  r  varies  from  0  to  2.  The  second  integration  is 
therefore  - 


VOLUME  281 

Finally,  0  must  vary  from  0  to  2  tt,  and  the  third  integration  is 


If  we  put  together  what  we  have  done,  we  have 

/I  2  JT    /» 2     /."v  5  —  r*  9  TT  /  y—  N 

F=/        (I  rrf^c/rcf2  =  =-(5V5-4). 

Jo        ./O    ./r2  3 

£X£RCIS£S 

1 .  Find  the  volume  bounded  by  the  paraboloid  z  =  x^  -\-  y^  and 
the  planes  x  =  0,  y  =  0,  and  z  =  4. 

Z  CC  7/^ 

2.  Prove  that  the  volume  bounded  by  the  surface  -  =  —  +  f- 

c      a^      ¥ 

and  the  plane  «  =  c  is  one  half  the  product  of  the  area  of  the  base 
by  the  altitude. 

3.  Find  the  volume  bounded  by  the  plane  z  —  0  and  the  cylinders 
x^  -\-  y'^=  V?  and  y'^=  c?  —  az. 

4.  Find  the  volume  cut  from  the  sphere  r^-\-z'^  =  a?  by  the  cylinder 
r  =  a  cos  B. 

5.  Find  the  volume  bounded  below  by  the  paraboloid  r^^az  and 
above  by  the  sphere  r^  -{-  z^  —  2  az  =  Q. 

6.  Find  the  volume  bounded  by  the  plane  XOY,  the  cylinder 
ic^  -j-  2/^  —  2  ax  =  0,  and  the  right  circular  cone  having  its  vertex  at  Oy 
its  axis  coincident  with  OZ,  and  its  vertical  angle  equal  to  90°. 

7.  Find  the  volume  bounded  by  the  surfaces  r^  =&«,«  =  0,  and 
r  =  a  cos  0. 

8.  Find  the  volume  bounded  by  a  sphere  of  radius  a  and  a  right 
circular  cone,  the  axis  of  the  cone  coinciding  with  a  diameter  of  the 
sphere,  the  vertex  being  at  an  end  of  the  diameter,  and  the  vertical 
angle  of  the  cone  being  90° . 

9.  Find  the  volume  of  the  sphere  of  radius  a  and  with  its  center 
at  the  origin  of  coordinates,  included  in  the  cylinder  having  for  its 
base  one  loop  of  the  curve  r^  =  a^  cos  2  9. 

10.  Find  the  volume  of  the  paraboloid  x^-\-y'^  =  2z  cut  off  by  the 
plane  z  =^x  -^-1. 

11.  Find  the  volume   of  the  solid  bounded  by  the  paraboloid 
z  =  x^  -\-  y"^  and  the  plane  z  =  x. 


282  REPEATED  INTEGRATION 

93.  Center  of  gravity  of  a  solid.  The  center  of  gravity  of  a  solid 
has  three  coordinates,  ^,  ^,  2,  which  are  defined  by  the  equations 

I  xdm  j  ydm  I  zdm 

[dm  j  dm  t  dm 

where  dm  is  the  mass  of  one  of  the  elements  into  which  the 
solid  may  be  divided,  and  rr,  ?/,  and  z  are  the  coordinates  of  the 
point  at  which  the  element  dm  may  be  regarded  as  concentrated. 
The  derivation  of  these  formulas  is  the  same  as  that  in  §  85 
and  is  left  to  the  student. 

When  dm  is  expressed  in  terms  of  space  coordinates,  the 
integrals  become  triple  integrals,  and  the  limits  of  integration 
are  to  be  substituted  so  as  to  include  the  whole  solid. 

We  place  dm  —  pd  F,  where  p  is  the  density.  If  p  is  constant, 
it  may  be  placed  outside  the  integral  signs  and  canceled  from 
numerators  and  denominators.    Formulas  (1)  then  become 

Vx=  CxdV,      Vy=CydV,      Vz=CzdV.  (2) 

Ex.  Find  the  center  of  gravity  of  a  body  bounded  by  one  nappe  of  a 
right  circular  cone  of  vertical  angle  2  a  and  a  sphere  of  radius  a,  the  center 
of  the  sphere  being  at  the  vertex  of  the  cone. 

If  the  center  of  the  sphere  is  taken  as  the  origin  of  coordinates  and  the 
axis  of  the  cone  as  the  axis  of  z,  it  is  evident  from  the  symmetry  of  the 
solid  that  x  =  y  =  0.  To  find  z  we  shall  use  cylindrical  coordinates, 
the  equations  of  the  sphere  and  the  cone  being  respectively 

r^  •{■  z^  =  a?     and     r  =■  z  tan  a. 

As  in  Ex.  2,  §  92,  the  surfaces  intersect  in  the  circle  r  =  a  sin  a:  in 
the  plane  z  —  a  cos  a.    Therefore 

I  f  rdOdrdz  =  %  ira^(l  -  cos  a) 

0       Jo  Jrctna 

zdV=j        I  I  rzdddrdz  =  I  Tra"^  sin- a. 

Jo       Jo  Jrcttxa 

Therefore,  from  (2),  2  =  §  (1  +  cos  a). 


CENTER  OF  GRAVITY  283 

EXERCISES 

1.  Find  the  center  of  gravity  of  a  solid  bounded  by  the  paraboloid 

-  =  — ,  4-  7?  and  the  plane  z  ■=  c. 

2.  A  ring  is  cut  from  a  spherical  shell,  the  inner  radius  and  the 
outer  radius  of  which  are  respectively  4  ft.  and  5  ft.,  by  two  parallel 
planes  on  the  same  side  of  the  center  of  the  shell  and  distant  1  ft. 
and  3  ft.  respectively  from  the  center.  Find  the  center  of  gravity  of 
this  ring. 

3.  Find  the  center  of  gravity  of  a  solid  in  the  form  of  the  frustum 
of  a  right  circular  cone  the  height  of  which  is  A,  and  the  radius 
of  the  upper  base  and  the  radius  of  the  lower  base  of  which  are 
respectively  v^  and  r^. 

4.  Find  the  center  of  gravity  of  that  portion  of  the  solid  of 
Ex.  2,  p.  73,  which  is  above  the  plane  determined  by  OA  and 
OB  (Fig.  31). 

5.  Find  the  center  of  gravity  of  a  body  in  the  form  of  an  octant 

of  the  ellipsoid  -^  +  75  +  — ,  =  1. 
^         (V-      If      c^ 

6.  Find  the  center  of  gravity  of  a  solid  bounded  below  by  the 
paraboloid  az  =  7^  and  above  by  the  right  circular  cone  z  -\-  r  =  2a. 

7.  Find  the  center  of  gravity  of  a  solid  bounded  below  by  the 
cone  z  =  r  and  above  by  the  sphere  r^  -{-  z^  =  1. 

8.  Find  the  center  of  gravity  of  a  solid  bounded  by  the  surfaces 

z  =  0,  7^ -\-  z^  =  1/,  and  r  =  a{a<  b). 

94.  Moment  of  inertia  of  a  solid.  If  a  solid  body  is  divided 
into  elements  of  volume  c?F,  then,  as  in  §  88,  the  moment  of 
inertia  of  the  solid  about  any  axis  is 

/=  CR^pdV=^p  CR^dV,  (1) 

where  E  is  the  distance  of  any  point  of  the  element  from  the 
axis,  and  p  is  the  density  of  the  solid,  which  we  have  assumed 
to  be  constant  and  therefore  have  been  able  to  take  out  of  the 
integral  sign.  If  M  is  the  total  mass  of  the  solid,  p  may  be 
determined  from  the  formula  M=pV, 


284  REPEATED  INTEGRATION 

If  the  moment  of  inertia  about  OZ,  which  we  shall  call  /^,  is 
required,  then  in  cylindrical  coordinates  R  =  r  and  dV=  rdOdrdz, 
so  that  (1)  becomes 

I,  =  p  fCCr^dddrdz.  (2) 

If  we  use  Cartesian  coordinates  to  determine  /^,  we  have 
R^—^-^-'if-  and  dV=  dxdydz^  so  that 

Similarly,  if  7^  and  I^  are  the  moments  of  inertia  about  OY 
and  OX  respectively,  we  have 

/,  =  pfff(^^'-^  ^')  ^^^^^^'      ^.  =  ^jjj^^'^  ^'^  dxdydz,   (4) 

In  evaluating  (2)  it  is  sometimes  convenient  to  integrate 
with  respect  to  z  last.    We  indicate  this  by  the  formula 


='HI 


r'dddr,  (5) 


But  j  I  r^dOdr  is,  by  §  88,  the  polar  moment  of  inertia  of  a 

plane  section  perpendicular  to  OZ  about  the  point  in  which  OZ 
intersects  the  plane  section.  Consequently,  if  this  polar  moment 
is  known,  the  evaluation  of  (5)  reduces  to  a  single  integration. 
This  has  already  been  illustrated  in  the  case  of  solids  of  revolution. 

A  similar  result  is  obtained  by  considering  (3).  In  fact,  the 
ease  with  which  a  moment  of  inertia  is  found  depends  upon  a 
proper  choice  of  Cartesian  or  cylindrical  coordinates  and,  after 
that  choice  has  been  made,  upon  the  order  in  which  the  integra- 
tions are  carried  out. 

Equation  (3)  may  be  written  in  the  form 

I^=  p  III  x^dxdydz  +  P  I  I  I  y^dxdydz,  (6) 

and  the  order  of  integration  in  the  two  integrals  need  not  be 
the  same.    Similar  forms  are  derived  from  (4). 

The  theorem  of  §  89  holds  for  solids.  This  is  easily  proved 
by  the  same  methods  used  in  that  section. 


MOMENT  OF  INERTIA  285 

Ex.  Find  the  moment  of  inertia  about  OZ  of  a  cylindrical  solid  of 
altitude  h  whose  base  is  one  loop  of  the  curve  r  =  a  sin  3  d. 

The  base  of  this  cylinder  is  shown  in  Fig.  65,  p.  144.  We  have,  from 
formula  (2),  t     „.,„,a     , 

\    y  /»  _    /»  a  sin  3  0    nh 

where  the  limits  are  obtained  as  follows: 

First,  holding  r,  B,  d9,  dr  constant,  we  allow  z  to  vary  from  the  lower 
base  2  =  0  to  the  upper  base  z  =  h,  and  integrate.  The  result  pht^dBdr  is 
the  moment  of  inertia  of  a  column  such  as  is  shown  in  Fig.  127.  We 
next  hold  6  and  d$  constant  and  allow  r  to  vary  from  its  value  at  the 
origin  to  its  value  on  the  curve  r  =  a  sin  3  6,  and  integrate.  The  result 
^pAa*sin^3  Odd  is  the  moment  of  inertia  of  a  slice  as  shown  in  Fig.  127. 
Finally,  we  sum  all  these  slices  while  allowing  6  to  vary  from  its  smallest 

value  0  to  its  largest  value  -  •    The  result  is  —  ph  a%. 
3  32 

The  volume  of  the  cylinder  may  be  computed  from  the  formula 

V  =  j^J  J    rdedrdz=  j\  AaV. 

Therefore  M  =  ^\^  ph  a%     and     I^  =  |  Afa^. 

EXERCISES 

1.  Find  the  moment  of  inertia  of  a  rectangular  parallelepiped 
about  an  axis  through  its  center  parallel  to  one  of  its  edges. 

2.  Find  the  moment  of  inertia  about  OZ  of  a  solid  bounded  by 
the  surface  z  =  2  and  z"^  =  r. 

3.  Find  the  moment  of  inertia  of  a  right  circular  cone  of  radius  a 
and  height  h  about  any  diameter  of  its  base  as  an  axis. 

4.  Find  the  moment  of  inertia  about  OZ  of  a  solid  bounded  by 
z  _x'^  .  ?/^ 
c      a^ 

5.  Find  the  moment  of  inertia  of  a  right  circular  cone  of  height  h 
and  radius  a  about  an  axis  perpendicular  to  the  axis  of  the  cone  at 
its  vertex. 

6.  Find  the  moment  of  inertia  of  a  right  circular  cylinder  of 
height  h  and  radius  a  about  a  diameter  of  its  base. 

7.  Find  the  moment  of  inertia  about  OZ  of  the  portion  of  the 
sphere  r^-\-  z^=  a}  cut  out  by  the  plane  «  =  0  and  the  cylinder 
r  —  a  cos  0. 


the  paraboloid  ~  =  -^  +  ^  and  the  plane  ^  =  c. 


286  REPEATED  INTEGRATION 

8.  Find  the  moment  of  inertia  about  OX  of  a  solid  bounded  by 
the  paraboloid  z  —  r^  and  the  plane  z  =  2. 

9.  Find  the  moment  of  inertia  about  its  axis  of  a  right  elliptic 
cylinder  of  height  ^,  the  major  and  the  minor  axis  of  its  base  being 
respectively  2  a  and  2  h. 

10.  Find    the    moment   of   inertia    about    OZ   of    the    ellipsoid 

GENERAL  EXERCISES 

1.  Find  the  center  of  gravity  of  the  arc  of  the  curve  x^  ■\-y'^  =  a^, 
which  is  above  the  axis  of  x. 

2.  A  wire  is  bent  into  a  curve  of  the  form  9?/^  =  x^.  Find  the 
center  of  gravity  of  the  portion  of  the  wire  between  the  points  for 
which  X  =  0  and  x  =  5  respectively. 

3.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 
ay^  =  x^  and  any  double  ordinate. 

4.  Find  the  center  of  gravity  of  the  area  bounded  by  the  axis 
of  X,  the  axis  of  ^/,  and  the  curve  i/^  =  8  —  2  a:-. 

5.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curves 
2/  =  x^  and  i/  =  —^  the  axis  of  x,  and  the  line  x  =  2. 

6.  Find  the  center  of  gravity  of  the  area  bounded  by  the  axes 
of  X  and  ?/  and  the  curve  x  =  a  GOS^<f>,  1/  =  a  sin^^. 

7.  Find  the  center  of  gravity  of  the  area  bounded  by  the  ellipse 

x^      ip' 

-^  +  y^  =  1  (a>  h)^  the  circle  x'^  -{-  tf  =  a^,  and  the  axis  of  y. 

8.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 
x''=%y  and  the  circle  ic^  +  t/^  =  128. 

9.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curves 
x^  —  a  (y  —  b)  =  0,  x^  —  ay  =  0,  the  axis  of  y,  and  the  line  x  =  c. 

10.  Find  the  center  of  gravity  of  an  area  in  the  fqrm  of  a  semi- 
circle of  radius  a  surmounted  by  an  equilateral  triangle  having  one 
of  its  sides  coinciding  with  the  diameter  of  the  semicircle. 

11.  Find  the  center  of  gravity  of  an  area  in  the  form  of  a  rec- 
tangle of  dimensions  a  and  b  surmounted  by  an  equilateral  triangle 
one  side  of  which  coincides  with  one  side  of  the  rectangle  which  is 
b  units  long. 


GENERAL  EXERCISES  287 

12.  Find  the  center  of  gravity  of  the  segment  of  a  circle  of 
radius  a  cut  off  by  a  straight  line  h  units  from  the  center. 

13.  From  a  rectangle  h  units  long  and  a  units  broad  a  semicircle 
of  diameter  a  units  long  is  cut,  the  diameter  of  the  semicircle 
coinciding  with  a  side  of  the  rectangle.  Find  the  center  of  gravity 
of  the  portion  of  the  rectangle  left. 

14.  Find  the  center  of  gravity  of  a  plate  in  the  form  of  one  half  of 
a  circular  ring  the  inner  and  the  outer  radii  of  which  are  respectively 
r^  and  r^. 

15.  In  the  result  of  Ex.  14,  place  r^=  r^-\-  Ar  and  find  the  limit  as 
Ar— vO,  thus  obtaining  the  center  of  gravity  of  a  semicircum  fere  nee. 

16.  Find  the  center  of  gravity  of  a  plate  in  the  form  of  a  T-square 
10  in.  across  the  top  and  12  in.  tall,  the  width  of  the  upright  and 
that  of  the  top  being  each  2  in. 

17.  From  a  plate  in  the  form  of  a  regular  hexagon  5  in.  on  a  side, 
one  of  the  six  equilateral  triangles  into  which  it  may  be  divided  is 
removed.    Find  the  center  of  gravity  of  the  portion  left. 

18.  Find  the  center  of  gravity  of  a  plate,  in  the  form  of  the  ellipse 

-^  +  7^  =  1  (a  >  J),  in  which  there  is  a  circular  hole  of  radius  c, 
(V-      If         ^  ■'' 

the  center  of  the  hole  being  on  the  major  axis  of  the  ellipse  at  a 
distance  d  from  its  center. 

19.  Find  the  center  of  gravity  of  the  solid  formed  by  revolving 

x^      if 
about  OY  the  surface  bounded  by  the  hyperbola  "i  —  7^  =1  and  the 

lines  y  =  0  and  y  =  b. 

20.  Find  the  center  of  gravity  of  the  solid  generated  by  revolving 
about  the  line  x  =  a  the  area  bounded  by  that  line,  the  axis  of  a-,  and 
the  parabola  y"^  =  kx. 

21.  Find  the  center  of  gravity  of  the  segment  cut  from  a  sphere  of 
radius  a  by  two  parallel  planes  distant  respectively  h^  and  h^  (h^  >  h^ 
from  the  center  of  the  sphere. 

22.  Find  the  moment  of  inertia  of  a  plane  triangle  of  altitude  a  and 
base  b  about  an  axis  passing  through  its  center  of  gravity  parallel  to 
the  base. 

23.  Find  the  moment  of  inertia  of  a  parallelogram  of  altitude  a 
and  base  b  about  its  base  as  an  axis. 


288  REPEATED  INTEGRATION 

24.  Find  the  moment  of  inertia  of  a  plane  circular  ring,  the  inner 
radius  and  the  outer  radius  of  which  are  respectively  3  in.  and  5  in., 
about  a  diameter  of  the  ring  as  an  axis. 

25.  A  square  plate  10  in.  on  a  side  has  a  square  hole  5  in.  on  a 
side  cut  in  it,  the  center  of  the  hole  being  at  the  center  of  the  plate 
and  its  sides  parallel  to  the  sides  of  the  plate.  Find  the  moment  of 
inertia  of  the  plate  about  a  line  through  its  center  parallel  to  one 
side  as  an  axis. 

26.  Find  the  moment  of  inertia  of  the  plate  of  Ex.  25  about  one  of 
the  outer  sides  as  an  axis. 

27.  Find  the  moment  of  inertia  of  the  plate  of  Ex.  25  about  one 
side  of  the  hole  as  an  axis. 

28.  Find  the  moment  of  inertia  of  the  plate  of  Ex.  25  about  one 
of  its  diagonals  as  an  axis. 

29.  A  square  plate  8  in.  on  a  side  has  a  circular  hole  4  in.  in 
diameter  cut  in  it,  the  center  of  the  hole  coinciding  with  the  center 
of  the  square.  Find  the  moment  of  inertia  of  the  plate  about  a  line 
passing  through  its  center  parallel  to  one  side  as  an  axis. 

30.  Find  the  moment  of  inertia  of  the  plate  of  Ex.  29  about  a 
diagonal  of  the  square  as  an  axis. 

31.  Find  the  moment  of  inertia  of  a  semicircle  about  a  tangent 
parallel  to  its  diameter  as  an  axis. 

32.  Find  the  polar  moment  of  inertia  of  the  plate  of  Ex.  25  about 
its  center. 

33.  Find  the  polar  moment  of  inertia  of  the  entire  area  bounded 
by  the  curve  r^  =  c^  sin  3  Q  about  the  pole. 

34.  Find  the  polar  moment  of  inertia  of  the  area  bounded  by  the 
cardioid  r  =  a  (  1  +  cos  0')  about  the  pole. 

35.  Find  the  polar  moment  of  inertia  of  that  area  of  the  circle 
r  =  a  which  is  not  included  in  the  curve  r  =  a  sin  2  B  about  the 
pole. 

36.  Find  the  moment  of  inertia  about  OF  of  a  solid  bounded  by 
the  surface  generated  by  revolving  about  OF  the  area  bounded  by  the 
curve  y^=Xy  the  axis  of  y,  and  the  line  y  =  2. 

37.  A  solid  is  in  the  form  of  a  hemispherical  shell  the  inner 
radius  and  the  outer  radius  of  which  are  respectively  r^  and  r^.  Find 
its  moment  of  inertia  about  any  diameter  of  the  base  of  the  shell  as 
an  axis. 


GENERAL  EXERCISES  289 

38.  A  solid  is  in  the  form  of  a  spherical  cone  cut  from  a  sphere 
of  radius  a,  the  vertical  angle  of  the  cone  being  90°.  Find  its 
moment  of  inertia  about  its  axis. 

39.  A  solid  is  cut  from  a  hemisphere  of  radius  5  in.  by  a  right 
circular  cylinder  of  radius  3  in.,  the  axis  of  the  cylinder  being 
perpendicular  to  the  base  of  the  hemisphere  at  its  center.  Find  its 
moment  of  inertia  about  the  axis  of  the  cylinder  as  an  axis. 

40.  An  anchor  ring  of  mass  M  is  bounded  by  the  surface  generated 
by  revolving  a  circle  of  radius  a  about  an  axis  in  its  plane  distant 
b(b  >  a)  from  its  center.  Find  the  moment  of  inertia  of  this  anchor 
ring  about  its  axis. 

41.  Find  the  moment  of  inertia  of  the  elliptic  cylinder  —  +  7-  =1 

(a  >  ^),  its  height  being  7i,  about  the  major  axis  of  its  base. 

42.  Find  the  center  of  gravity  of  the  solid  bounded  by  the  cylinder 
r  =2a  cos  6,  the  cone  z  =  r,  and  the  plane  ^  =  0. 

43.  Find  the  moment  of  inertia  about  OZ  of  the  solid  of  Ex.  42. 

44.  Find  the  volume  of  the  cylinder  having  for  its  base  one  loop 
of  the  curve  r  —  2a  cos  2  6,  between  the  cone  z  =  2r  and  the  plane 
z  =  0. 

45.  Find  the  center  of  gravity  of  the  solid  of  Ex.  44. 

46.  Find  the  moment  of  inertia  about  OZ  of  the  solid  of  Ex.  44. 

47.  Find  the  volume  of  the  cylinder  having  for  its  base  one  loop 
of  the  curve  r  =  a  cos  2  B  and  bounded  by  the  planes  ^  =  0  and 
z  =  X  -{-  2  a. 

48.  Find  the  moment  of  inertia  about  OZ  of  the  solid  of  Ex.  47. 

49.  Find  the  volume  of  the  cylinder  r  =  2a  cos  0  included  between 
the  planes  z  =  0  and  z  =  2x  -\-  a. 

50.  Find  the  moment  of  inertia  about  OZ  of  the  solid  of  Ex.  49. 

51.  Through  a  spherical  shell  of  which  the  inner  radius  and  the 
outer  radius  are  respectively  r^  and  r^,  a  circular  hole  of  radius 
a{a  <  r^  is  bored,  the  axis  of  the  hole  coinciding  with  a  diameter 
of  the  shell.  Find  the  moment  of  inertia  of  the  ring  thus  formed 
about  the  axis  of  the  hole. 


ANSWERS 


[The  answers  to  some  problems  are  intentionally  omitted.] 
CHAPTER  I 


Page  4  (§  2) 

1.  21|f.  4.  106 ft.  per  second. 

2.  1^|.  5.  33.97  mi.  per  hour  for  entire  trip.  8.  1^  mi.  per  hour. 

3.  iO\^.  6.  1.26.  -    --  . 


7.  21. 


9.  98.4. 


Page  5  (§  2) 
10.  196.8. 

Page  7  (§  3) 

1.  96  ft.  per  second. 

Page  8  (§  3) 

3.  128  ft.  per  second. 

4.  74  ft.  per  second. 


2.  128  ft.  per  second. 


5.  68  ft.  per  second. 

6.  52  ft.  per  second. 


Page  11  (§  5) 

1.  12«i2;  24«i.         3.  85;  32;  6.         6.  5,  4,  when  «  =  2  ;   10,  6,  when  i  =  3. 

2.  16;  14.  4.  42;  57.  S.  Sat^  +  2bt  +  c  ;  Qat  +  2b. 


Page  13  (§  6) 

1.  —  sq.  ft.  per  second. 

TT 

4.  4  7rr2. 

Page  14  (§  6) 

6.  8  7rr.  7.  3(edge)2. 


27r 


3.  8  Trr^  cu.  in.  per  second. 


6.  16  Trrsq.  in.  per  second. 


8.  67rr2 


9.  18. 


10.  2 


Page  18  (§  7) 

1.  8x.  2.  3x2  +  4x. 

6.    -        ' 


(2  +  x)2 


CHAPTER  II 

3.  4x3-2x. 

7.  x^  +  x+1. 
291 


--!. 


6.  2x 

x8 


8.3--^, 


292  ANSWERS 

Page  21  (§  9) 

1.  Increasing  if  x  >  2  ;  decreasing  if  a;  <  2. 

2.  Increasing  if  x  >  —  ^  ;  decreasing  if  a;  <  —  ^. 

3.  Increasing  if  x  <  |  ;  decreasing  if  x  >  ^. 

4.  Increasing  if  x  <  —  ^  ;  decreasing  if  x  >  —  |^. 

6.  Increasing  ifx<  —  2orx>l;  decreasing  if  —  2  <  x  <  1. 

6.  Increasing  ifx<— 5orx>3;  decreasing  if  —  5  <  x  <  3. 

7.  Increasing  ifx<  —  lorx>f;  decreasing  if  —  1  <  x  <  ^. 

8.  Always  increasing. 

9.  Increasing  ifx<  —  lor—  ^<x<l;  decreasing  if  —  l<x<—  Jorx>l. 
10.  Increasing  if  x  >  1 ;  decreasing  if  x  <  1. 

Page  24  (§  10) 

1.  When  t<—lort>l;  when  —l<t<l. 

2.  When  i  <  5  ;  when  t  >  5. 

3.  Wlien  t<2ort>4;  when  2  <t<4. 

4.  Always  moves  in  direction  in  which  s  is  measured. 
6.  When  t>  ^;  when  t  <  |. 

6.  Always  increasing. 

7.  Always  decreasing. 

8.  Increasing  when  t>  2  ;  decreasing  when  t  <2. 

9.  Increasing  when  t>  ^  ;  decreasing  when  t  <  ^. 
10.  Increasing  when  t  <  §  ;  decreasing  when  i  >  §. 

Page  26  (§  11) 

1.  ^  Trh'^.  2.  6  7r/i  sq.  ft.  per  second. 

Page  27  (§  11) 

3.  0.2  cm.  per  second.  5.  0.26.  ^-(tt\h'  ht\^ 

4.  20.9  sq.  in.  per  second.      6.  64  cu.  ft.  per  second.        '8  &    ^  • 

8.  4  TT  (i2  +  12 «  +  36) ;  t  is  thickness. 
Page  31  (§  13) 

1.  1.46.  3.  0.46;  2.05.  6.  2.41. 

2.  -2.07.  4.  1.12;  3.93.  6.  -2.52. 

Page  35  (§  14) 

1.  Sx-y-9  =  0.  6.x +  2/ +  1  =  0.  *12.  tan-i^. 

2.  2 X  +  3 y  +  3  =  0.  7.  X  +  2 2/  +  8  =  0.  13.  tan-il2. 

3.  21  X- 22/ -13  =  0.  8.  4x-3?/-l=0.  H.  |. 

4.2/4-3  =  0.  _  9.  12x-42/-5  =  0.  U.  2x-3y-16  =  0. 

5.  V3x-2/-2V3-2  =  0.  10.  5x  -  Oy  -  4  =  0.  16.  (-l^^^,  21*3). 

*The  symbol  tan-^  I  represents  the  angle  whose  tangent  is  ^  (cf.  §  46). 


ANSWERS 

29: 

Page  39  (§  15) 

1.  (-  §,  2|). 

1,  2x-y-l  =  0. 

2.  i'l  Hh 

8.  4x  +  y  +  4  =  0. 

8.  (0,  4),  (2,  0). 

9.  (-  3,  10),  (1,  2). 

4.  (1,  7),  (3,  3). 

10.  27x  +  272/-86  = 

z0;x  +  y-2  =  0. 

6.  (-  2,  0),  (1,  -  9). 

11.  18  X- 27  2/ +  80  = 

=  0;  18x-27y-28  =  0 

6.  (-  1,  -  3),  (3,  29). 

12.  tan-i§. 

Page  43  (§  17) 

1.  25sq.in. 

3.  6  ft. 

^    ina^VS 

2.  Lerxgth  is  twice  breadth. 

4.  50. 

^'         9 

Page  44  (§  17) 

6.  Depth  is  one-half  side  of  base. 

7.  2  portions  4  ft.  long ;  4  portions  1  ft.  long. 

2aV6 


8.  Breadth  = ;  depth  = 

^;base=^ 
4     '  4 

;  2547  cu.  in. 


9.  Altitude  = 

10.  2000  cu.  in  . 

11.  Height  of  rectangle 

5 

12.  -7=in. 
V3 


(p  =  perimeter) 


radius  of  semicircle  ;  semicircle  of  radius, 


Page  46  (§  18) 

1.  426  ft. 


Page  47  (§  18) 

2.  46  ft. 

3.  95  ft. 

4.  38|ft. 


5.  576  ft. 

6.  2/  =  x2  +  3  a;  -  6. 

7.  y  =  2x^-\-x^-ix  +  6. 


621. 


Page  49  (§  19) 
1.  7^.       2.  13^. 

Page  53  (§  20) 

8.  0.0001  ;  0.000001  ;  0.00000001. 

9.  0.000009001  ;  0.000000090001. 

Page  54  (§  21) 

1.  72  sq.  in. 

„     57r  .  TT  . 

2.  —  cu.  m. ;  -  sq.  in. 

16  '  2    ^ 

«    27r  37r 

3.  — cu.  in. ;  —  cu.in. 

5  25 


8.  y  =  7+4a;-|x2-  i  xs. 

9.  y=l(x2-33). 

10.  2/=^x3-  ^x2  +  x+  |. 


4.  36.       6.  IJ.       6.  lj%.       7.  2^.       8.  2^. 


10.  0.000003  sq.  in. 

11.  456.58  cu.in. 


4.  27.0054  cu.  in. 
6.  28.2749  cu.  in. 

6.  606.0912. 

7.  0.0012. 

8.  6.99934. 


294 

Page  55  (General  Exercises) 

5  ^  2x 


(1  -  X)2  (X2  -f  1)2 

2a  ^         4x 

4. 


ANSWERS 

.      6.     1    . 

2Vx 

7.  --^ 

Vx2  +  1 

6             '     . 

13.  Increasing  if  x  >  —  2  ; 

2Va;3 

decreasing  if  x  <  —  2. 

(a  -  x)2  (x2  +  1)2 

14.  Increasing  if  —  |<x<§orx>2;  decreasing  if  x  <  —  |  or  |  <  x  <  2. 

16.  Increasing  if  x  > ;  decreasing  if  x  < 

2a  2a 

16.  Increasing  if  x  < :^  or  x  >  — r: ;  decreasing  if <x  <  — =:  • 

V3  VS  V3  .         VS 

17.  Increase  if  x  < —  ;  decrease  if  x  >  — -• 

3  3 

Page  56  (General  Exercises) 

18.  1<  «  <  5.  ,         19.  2  <  i  <  5  ;  41 

20.  Up  when  0  <  f  <  61 ;  down  when  6\  <t<  121. 

21.  Increasing  when  t  >  4  ;  decreasing  when  i  <  4. 

22.  V  increasing  when  i  <  3,  u  decreasing  when  ^  >  3  ;  speed  increasing  when 

2  <  i  <  3  or  f  >  4,  speed  decreasing  when  i  <  2  or  3  <  i  <  4. 

23.  Increasing  when  l<£<2ori>3;  decreasing  when  i  <  1  or  2  <  i  <  3. 

24.  0.0055  in.  per  minute. 

25.  8.6  in.  per  second.  27.  x-\-  2y  +  6  =  0. 

26.  1  sq.in.  per  minute.  28.  7x  +  5?/  +  1  =  0. 

Page  57  (General  Exercises) 

29.  X- 2  =  0.  31.  2x-?/  +  3  =  0.  33.  (-^,  3|). 

ZO.  x-2y-7  =  0.  32.  tan-i  |.  34.  (11,  0). 

35.  (2,  -2).  41.  102. 

36.  (-  1,  13),  (5,  -  95).  42.  tan-i  j%. 

37.  (-  3,  13),  (1,  -  19).  43.  X  -  ?/  -  11  =  0. 

38.  (-4,20).  44.  (1,  -1),(-^,  -,V\). 

Page  58  (General  Exercises) 

46.  62  ft.  long.  52.  ?/  =  x2  +  3x  -  13. 

47.  Altitude  of  cone  is  |  radius  of  sphere.  63.  y  =  ^  x^  —  x^  +  7x. 

AQ    AU-.  ^         *l4^      -A      ,u  '1'^^  64.85^. 

48.  Altitude  =  \  ——;  side  of  base  =  A -_    ooi 

\243  \  27  56.  28|. 

49.  2  pieces  3  in.  long  ;  3  pieces  1  in.  long.  56.  20^. 
60.  600  ft.  67.  72. 

51.  56  ft.  '  69.  0.0003. 

Page  59  (General  Exercises) 

60.  0.00629.  64.  0.09  cu.  in.  67.  24.0024  sq.  in. 

62.  288  TT  cu.  in.        66.  0.0003.  68.  0.4698. 

63.  161.16  cu.  in.       66.  354.1028  ;  353.8972. 


ANSWERS 

295 

CHAPTER  III 
Page  66  (§  23) 

1.  3,^.             3.  52^15.         6.  5.;^.             7.  36^i. 

2.  23^.           4.  166§.          6.  42§.           8.  21. 

9.  96.              11.  42§. 
10.  10§.           12.  10§. 

Page  67  (§  24) 

1.  150  ft.                          2.  140  ft. 

3.  57^  ft. 

Page  68  (§  24) 

4.  When  ^  <  «  < -i^  ;  83^  ft.                          6. 

8000  TT  ft.-lb. 

Page  70  (§  25) 

1.  8^T.                         2.  21 T.                         3.  3T. 

4.  l^T. 

Page  71  (§  25) 

6.  Approx.  24131b.                             7.  585  j^T. 
6.  4^\T.;  4^\T.                                8.  117, \T. 

11.  2.1  ft.  from  upper  side. 

9.  234  §  T. 
10.  2/^  T. 

Page  75  (§  26) 

1.  21  ^TT.                       3.  34^2^  TT.                   6.  3331  c'u 
625  V3                     4.  l^TT.                   t.S^^TT. 
4      '                   6.  557f7r.                    8.  2i. 

L.  in.                   9.  25f  TT. 

10.  213^ 

11.  761". 

Page  76  (General  Exercises) 

1.  6|ft.                             5.  20.                                     8. 

2.  81ft.                                  4a2V2                             9. 

3.  lO^ft.                                     3       '                          12. 

4.  8l|mi.                         7.  8. 

21tV. 

,  Reduced  to  |  original 
pressure. 

Page  77  (General  Exercises) 

13.  Twice  as  great.                   17.  6if7r. 

14.  l^T.                                       ^^    144  7rV3 

15.  mw.                                 ^^'       35      ■ 

16.  68T^7r.                                 19.  96  TT. 

20.  341^  cu.  in. 

21    32  V3 

3 
23.  (a/i2-i;i8)7r. 

Page  78  (General  Exercises) 

24.  8  7r.                26.  1151.                26.  34^^^^  tt. 

29.  728,949  ft.-lb.                                        30. 

27.  9.              28.  204 1. 
5301  ft.-lb. 

CHAPTER  IV 
Page  81  (§  28) 

1.  x2  +  ?/2_8x-l-4i/  +  ll  =  0. 

2.  x^  +  y^  +  2y-24  =  0. 

6.  3x- 22/  -1-  4  =  0. 

3.  (-3,5);  5. 

4.  (-1,^-);  2. 

296  ANSWERS 

Page  84  (§  30) 

1.  (-2,0).  .  4.(0,-13). 

2.  (0,  1).  6.  SUft.  ^    IOttVc. 

3.  (1^,0).  6.  lOVlOft.  8 

Page  85  (§  30) 

Q.y^  +  6x-9  =  0.  10.  x2-4x-12i/  +  16  =  0. 

Page  87  (§  31) 

l.(..0);(.V7,0);^.  3.(.f,0>(4«,0);l. 

V5  .  /    V2   ^\ .  /    Ve   .\   Vs 

6.  9x2  4.  25^2  _  36a,  _  139  =  0. 
6.  49x2  +  24 1/2  -  120?/ -  144  =  0. 


2.(0,±3);(0,±V6);4l  4.(.f,o);(±-f,o); 


Page  91  (§  32)  _ 

1.  (±3,0);  (±Vl3,  0);  2x±3y  =  0',  -^. 

2.  (±2,  0);  (±  Vl3,  0);  3x±22/  =  0;  — .    _ 

3.  (0,  ±  V2);  (0,  ±  V5);  V2x±V3i/  =  0;  ^. 

4.  (±  2  V2,  0)  ;  (±  4,  0) ;  X  ±  2/  =  0  ;  V2. 

6.  (0,  ±  IV  (^0,  ±  -^^;  X  ±  2y  =  0  ;  Vs. 


3x2  +  8a; 


7.    3x2-2/2 

-12x 

+  9  =  0. 

8.  3x2-1/2 

+  4?/ 

-  16  =  0. 

Page  102  (§  36) 

1. 

18x2  +  22x-3. 

2. 

4x2(x  +  3). 

3. 

6x(x4-l). 

4 

36  X 

(X2  -  9)2 

6. 

x2  +  4  X  +  1 

(X2  +  X  +  1)2 

6. 

3\Vx      VxV 

7. 

2.-1+4-4- 

2Vx3  + 4x2  +  1 
10.  3xVx2  +  4. 


11.  -  3xVa2-x2. 

X 


13. 


14. 


V(9  -  X2)3 

3x(x8  +  4) 
V(x6  + 10x2 +  3)2 

1 


8.  2(8x  +  3)(4x2  +  3x  +  l).  (x-l)Vx2-l 


16. 

2(x-6)V(a;- 

16.^-^^^ 

-a)(x-6) 

V9-X2 

^^     2.X2  +  X-1 
Vx2-1 

ige  104  (§  37) 

J    ay-x2 

4. 

ANSWERS  297 

d-h      ^Q  «'  2j    (x  +  l)(3xHx+2) 


19. 


V(a2  +  X2)3  Vx2  +  1 

1  ^»  X8 


(X  +  1)  VX=^  -  1  V  (X2  +  9)8 

3  x2  +  6  „„  1 


V(x3  +  3)4  ^(1  +  xV 


•\Jy  +  X  —  V?/  —  X  _  x^  _  ^  2  a^x 

y^  -  ax  '  v^Tx  +  ^v-x  '     y'^'      y^ 

g  2xy  5_?^._A.  8.-^;^. 

x2  +  4a2"  ■       Sy'       3y^'  x^    2x^ 


3. 


y  .  4x.       16  9.  -^■ 


2*  '  9?/'       9?/3*  xi    Sx^y-s 


x  +  y{x  +  y) 

y  +  2,    2y  +  4  ^j        2x  +  ^.  6«^ 

*       x  +  3'(x  +  3)2'  •       x-{-2y'       {x-{-2yf' 

Page  105  (§  38) 

1.  3x-4  2/  +  2  =  0.       2.x- 7?/  + 5  =  0.        3.  (- 2,  -  1).      4.  tan-Uj^. 

Page  106  (§  38) 

TT  ^-19  ^0"  tan-13.  12.  tan-if ;  tan-if. 

8.  -.  9.  -;  tan      ^^.  ^^    tan-i|.  13.  tan-i^- 

Page  110  (§  40) 


29  2g 


1.  x8  =  8  ?/ ;  V4  +  9^4.  ^    (v^  sin  2  a       Uq^  sin2 a^ 

2.  x  =  (?/-l)2;  V4t2+i. 

3.  x2-6x  +  9?/  =  0;  |  Vll7- 48  i  +  16  ^2. 

4.  (3,  1).  9.  ~ ;  uo  ;  a. 

5.  {y  -  2)2  =  (X  +  3)3 ;  « V4  +  9  ^2.  ^9 

6.  x^  +  2/^  =  2 ;  8  Vl  -  2  t  +  2  £2.  10-  ;^- 

7  /V^f^^,  ^Ji^in^X  11.  y  =  xi2.na ^ 

\        2g  2g      }'  ^^o 


2vn  cos2  a 


Page  112  (§  41) 

1.  3V2ft.  per  second.         2.  12.5  ft.  per  second.         3.  ^^^  in.  per  minute. 

15 

4.  Circle;  —  ft.  per  second  (x  =  distance  of  point  from  wall). 

X 

5.  2.64  ft.  per  second. 

Page  113  (§41) 

6.  0.13  cm.  per  second.         7.  0.21  in.  per  minute.         8.  6.6  ft.  per  second. 

2  X    I    d  2/ 

9.  — - — -  ft.  per  second,  where  x  is  the  distance  of  top  of  ladder,  and  y  is 
2x+y 

distance  of  foot  of  ladder,  from  base  of  pyramid. 


298 


ANSWERS 


Page  114  (General  Exercises) 

20.  31x  +  8?/  +  9a  =  0. 

21.  x{~^x  +  Vi'^y  =  a^. 


28.  -:  tan- 17. 
2 


27. 

Page 

36. 
37. 

38. 


-;  tan-i- 
4'  3 

tan-if. 


39.  !r. 

''         12 

30.  tan-i  — 

5 


IT  1 

31.  -;  tan-1- 

2'  2 

32.  -;  tan-i- 
2  2 


34.  y 


8 ^  2V(f-^  +  l)*+4t2 


x'-^+  4 


33.  x*+ */^=  1;   3^ 
36.  2a^y  =  2abx-  gx^ ; 


115  (General  Exercises) 


(X  +  2)2  =  (y  -  1)3 ;    |V9i  +  4.       41.  5.8  mi.  per  hour;  28.8  mi 
20  ft.  per  second  ; 
10  VS  ft.  per  second  ;  (100,  20) 
0 


v^  =  ±^-^;  v,  =  T 

a  a 


43.  — z===^  ft.  per  second,   where 


40. 

Page 
46. 
47. 
48. 
49. 
50. 
51. 

Page 
67. 

60. 


Page 


Velocity  in  path 

1 


2ct 


Vax  +  a'^. 


116  (General  Exercises) 

0.08  ft.  per  minute. 
0.01  in.  per  minute. 
1^  sq.  in.  per  minute. 
0.04  in.  per  second. 
Length  is  twice  breadth. 
Other  sides  equal. 

117  (General  Exercises) 
2.64  in. 


68. 


8  mi.  from  point  on  bank  nearest  to  A 

bm  .         ,      J  bn  .    .         ^ 

a ::===  mi.  on  land  ;  — -z===  mi.  in  water. 


Vs2  _  400 
s  is  length  of  rope  from  man 
to  boat. 

44.  0.06  ft.  per  minute. 

45.  -g^  Q  ft.  per  minute. 

52.  Breadth,  9  in. ;  depth,  9V3in. 
63.  Length  =  |  breadth.' 

54.  Side  of  base,  10  ft. ; 
depth,  5  ft. 

55.  Depth  =  ^  side  of  base. 

56.  Radius,  3  in. ;  height,  3  in. 

1  ,^    aVd 

—-.  59.  — — . 

V2  3 

61.  4 II  mi.  travel  on  land. 


Vw2 


m" 


Vw2 


63.  l^f  hr. 
118  (General  Exercises) 


64.  V  100  mi.  per  hour. 


3a 


65.  Velocity  in  still  water  =  —  mi.  per  hour. 

66.  Base  =  aV3  ;  altitude  =  §  6. 


ge 

126  (§  44) 

CHAPTER 

V 

1. 

15  cos  5  X. 

5. 

sin2x. 

2. 
S. 

2sin2  2xcos2x. 

6. 

7. 

5sin2  5xcos3  6x. 

5sec2^tan5^. 
2           2 

4. 

—  5  sin  10  X. 

8. 

-  3csc33xctn3x. 

ANSWERS 


299 


Page  127  (§  44) 


9.  sin^ 


10.  —  CSC* 


11.  X  sin  -  . 

2 


12.  2  sec  X  (sec  x  +  tan  x)^. 
Page  129  (§  45) 


13.  2  cos  4  X. 

14.  9  tan* 3  X. 

15.  2sec2x(sec22x+  tan"'^2x). 

16.  sin3  2xcos2  2x. 

17.  —  §  cos2xcos2  32/. 

^. 

X 


18. 


2.  s  =  3  sin 


Trt 


3.  TT ;  5. 

4.  At  mean  point  of   motion  ;   at 

extreme  points  of  motion. 


Page  134  (§  47) 


Vl-9x2 

1 

X  Vx2  -  1 

1 

V6  X  -  x2 
3 

Vl2x-9x-^ 

1 

2  +  2  X  +  x2 

1 

6.  At  extreme  points  of  motion ;  at 
mean  point  of  motion. 

6.  2  V(s  -  8)  (5  -  s) ;  4(4 -s). 

7.  -TT. 

8.  10;  2  7r. 


13. 


(x-1)  Vx'-^  — 2x 


7. 
8 

2x 

x4  +  l 

1 

9. 
10 

xv'25x2-l 

1 

xV4x2-l 
2 

11 

x2+4 

1 

12. 

(x+l)Vx2  +  2x 

1 

X  \  X-  —  1 


14.  2  Vl  -  xK 
16 


15. 


16. 


17. 


(x2+4)« 


Vx2-4 


X 

2a 


x2  +  a2 


Vl-X2 


18.  cos-iVl  -x2. 


Page  136  (§  48) 

1.  Va:=±  9.42  ft.  per  second  ;  Vy  =  T  36.46  ft.  per  second. 

Page  137  (§  48) 

3.  3  radians  per  unit  of  time. 

Page  138  (§  49) 

„        6  sin  0  ,6 

7. —  ;  0  =  cos- 1  -  • 

a  —  b  cos  <t>  a 


2.  5.3. 


Page  141  (§  50) 


1.  §(x  +  3)Vx2  +  3x. 

2.  3(axy)i 


5V5 


5. 


itVit 


6.  atp. 


300 


ANSWERS 


Page  145  (§  51) 

17.  Origin  ;  ^Vs,  1^      18.  Origin  ;   ^±  a^?,  ^j 

20.0rigin;(±a,^);(±^,g;(i^_,^). 

21.  r2sin2^  =  8. 

22.  r  =  4a(cos^+  sin^). 

23.  r  =  2asin^. 

24.  r2  =  a2  cos  2  ^. 


19.  Origin 


(^•1) 


25.  X  —  a  =  0. 

26.  x^  +  2/2  -  2  ax  =  0. 

27.  x*  +  x22/2  =  a2?/2. 

28.  (X2  +  2/2)3  =  a2(x2-  2/2)2. 


Page  148  (§  52) 

1.  tan-i  ;\. 

Page  149  (General  Exercises) 

9.  2(l  +  sec22x). 

10.  sec*(3x  +  2). 

11.  cos2(2-3x). 
sec2  (x  —  2/)  +  sec2  (x  +  y) 
sec2  (x  —  y)—  sec2  (x  +  y) 

21.  - 


2.  TT- tan-i-§. 


3.  0. 


12. 


1  r 

17.  -Sin8-. 

2  4 

18.  tan4  2x. 

19.  ' 


13.  -3ctn2-csc*^- 

5         5 

14.  -  8csc24x(ctn4x  +  l). 

15.  a  tanaxsec2ax. 

16.  8cos3  2x  sin4x  cos6x. 

1 


20. 


(x  +  1)  Vx 
1 


2(x  +  l)Vx 

1 

xV49x2-l 

4 


V2  +  X  —  x2 
27.  


(x- 

■  i)V^ 

-2x 

95 

2 

V3- 

-4x- 

4x2 

26. 

2- 

-3x 

x2  +  4 


(x2-l)Vx2-2 

Page  150  (General  Exercises) 

29.  fcVa2sin2fci  +  b^  cos^  kt. 

30.  21^ 

31.  V4I;  7r  +  2tan-if. 
-.    16  V2 


V9- 
s  =  3;  2. 


x2 


35. 


(a2  +  62)i£ 
2a6V2 


7r3 
(7r2  +  1)1 
27r 
34.  2  a  Vs. 


33 


^  .  .       /      V6    7r\     /      V6    57r\ 
46.0ngm;(±-,-);(±-,-j. 

/^,tan-ilV 

Vv5       y 


47.  Origin  ; 


48.  Origin;  (^7^,  tan-i2V 


49.  Origin  ;  (2  a,  ^\  ;  ^2  a,  5^V 

Page  151  (General  Exercises) 

_  2asin2^  52.  (x2  +  y^  -  4  a^xy  =  0. 

"■  "     cos^     '  53.  (x2  +  2/2)2  +  2  ax(x2  +  ?/2)  -  a22/2  =  0. 

61.  r=actn^.  54.  tan-i§. 


ANSWERS 

i.  0;  tan-12. 

57.  '^ 
4 

1.  0;  -;  tan-isVs. 

58.? 
4 

301 

69.  IfiVsft. 

60.  72°. 

61.  V2  ft, 

62.  At  an  angle  tan-^A;  with  ground. 

Page  152  (General  Exercises) 

63.  12  in.  65.  a.  68.  15 sq.ft.;  10.94 sq.  ft.  per  second. 

64.  sVSft.      67.  6.1  ft.  per  second.      69.  26.7  mi.  per  minute. 

70.  (bsm6  -\ — )  times  angular  velocity  of  AB,  wliere  0  -  angle 

V  Va^-b^sin^e/ 

CAB. 

71.  ^^  ~  ^^"  +  ^-^  ~  ^^^  =  1 ;  V9sin^i4-4cos2'f ;  where  i  =  (2k-\-l)-. 

9  4  '  ^  '2 

72.  "i  _  L  =  1 ;  6  sec  3i  Vtan-'^ St +  4 sec^ 3 1. 
4       16        ' 

Page  153  (General  Exercises) 

73.  6  sin  20.  75.  tan-i2V2. 

74.  a  Vl  +  cos2  X ;   fastest  when  76.  tan-i|. 

x  =  k7r;  most  slowly  when  77.  0;  tan-isVs. 

x  =  {2k  +  l)-.  78.  tan-i^;  tan-i4V2. 

2  79.  tan-13;  tan-ij. 

CHAPTER  VI 
Page  162  (§  55) 

(The  student  is  not  expected  to  obtain  exactly  these  answers  ;  they  are  given 
merely  to  indicate  approximately  the  solution.) 

1.  ?/ =0.62  a; -0.76.  2.  Z=0.0017D. 

Page  163  (§  55) 

3.^=0.30(2.7)'.    4.  c  =  0,010(0.84)<.    5.  a =0.0000000048 /^.oe.    e.  pv^-^5  =  l0. 

Page  165  (§  56) 

12.  e-2^(3cos3ic-2sin3a;). 


1     1    -I 

^'  x^^   ''•  x2-9  13.  ctn-ia;. 

2.  ^(ex_e-a;).  g  1  14.  27x2  e8^. 

8.  2xa-^-ilna.  '  Vi^T4'  15.  5 e^- sin x. 

4.  „,sip-^x     ^"ct  g  _  2 

6,       2^  +  ^ 


V9x2  +  1 


x2  +  4a;_i  10.  —4  sec  2 X. 

2x  +  3  2(e2a:-e-2^) 

2  x2  +  6  X  +  9 '  ■     e2a:  +  e-2'f    *  xVx  +  1 


e^-f  e-^ 

17. 

2  sec^x. 

18. 

1 

302  ANSWERS 

Page  167  (§  57) 

a: 

1.  2/  =  5e2.  2.  ?/  =  45.22eO-oi^.  3.  ?/-7eO-847^.  4.  $739. 

Page  168  (§  57) 

6.  P  =  10000 eOo^^9'.  6.  c  =  0.01  e-o.0446<.  7,  2  min. 

Page  168  (General  Exercises) 

no.  p  =  0.018 1  +  24.  *11.  Load  =  192  -  6.6  length. 

Page  169  (General  Exercises) 

*12.  s  =  25(0.40)'.  *14.  t  =  OA  Vz.  _  *16.  y  =  OAOx^-^. 

*13.  c  =  0.010(0.83)'.  *15.  1=  0.023  V^. 


Page  170  (General  Exercises) 


*17.  pu=1620.  21.  2  CSC- 12a;.  26.  tan-ix. 

"•9^4*  22.  2(x+l)e-^-.  ^6.  0.898. 


19.  etna;.  no  2 


27.  15.8  hr. 


2  e2-^-  e^^  28.  1090 sec. 


e^  +  e-  ^  24.  a  tan^  ax.  29.  p  =  14. 7  e-o-oooo4  a. 

Page  171  (General  Exercises) 

31.  2V2e-2«;  2e-2^  33.  (e^^  +  ^)"\  35.  V2e'.  36.—. 

8  62^^  2 

3^   (a;2  +  l)t    3V3  gg    (1  +  e^)l 

X        '      2     '  '  n      ' 

2e2 

CHAPTER  VII 
Page  176  (§  59) 

.    ^  a;2       a;3  7.2       7.4       ,.6 

3        15        318 

4.x  +  L^  +  ll?.^,llU.5!+... 
2     32-4     6+2.4.6     7 


2      2    22^3    23^ 
9.  0.0872. 
10.  0.4695. 

*  Statement  in  regard  to  answers  to 

exercises  in  §  55  is  true  of  this  answer. 

ANSWERS  308 

Page  178  (§  60) 

l.ln5  +  -^      2         52       +3^^5^+        • 

■    2        2^6/        2    '       2!  2'       3!  "■■ 

4       2^  ^2         2!  2         3! 

W2  +  -L(x-l)+J__.(^-4..(^+'.... 
V2  2V2         2!    '       4V2         3! 

8.  0.7193.  9.  0.8480.  10.  3.0042. 

Page  179  (General  Exercises) 

/       a;2  ,  x3      X*      \  ^    ,      1    „      1  . 3  ^      1.3.5. 

''  -(^+  2  +  3  +I--7-  ''  ^-2"'  +  2— 4^^-274:6^^+  -•• 

2!        4!  6!  \         3        57/ 


,    1       V3 

1     X2 

'2*2! 

^    2      3!^ 

••• 

«    1^^^ 

1    x2 

V3   x8 

'•2+    2^- 

"2"2! 

2     3! 

*  *  • 

7.1   +   1x2  +  ; 

1.3   ^      1.3.5   « 
3.4^'+2.4.6^'  + 

... 

10. 

l-2x  +  2x2-2x8+  .... 

14.  0.9659. 

11. 

2x3      4x4 

16.  0.5150. 
16.  1.6487. 

12. 

2x2       2x3       4a;5 
^+2!    +3!          6..  +   -- 

17.  0.40546. 

18.  0.69315  ; 

1.0986. 

13. 

age 

x^  ,  x^  ,  3x«  , 
^+2  +  3  +  i0  +  -- 

180  (General  Exercises) 

19.  0.22314; 

20.  0.8473; 

1.6094. 
1.946. 

23. 

2.0305. 
2.9625. 

26,x-|  + 

x6 

^'      I 

24. 

5(2!) 

7(3!) 

25. 

^         x8           x6          a;7 
3(3!)      5(5!)      7(7!) 

28.X  +  I 

27.  a:  -  ^  + 

x3       X* 
3       4 

+  •.••. 

304  ANSWERS 

CHAPTER  VIII 
Page  183  (§  61) 

1.  3x2-8  x?/  +  82/2;  .  y  x 

0.    — 


—  4 aj2  +  16 x?/  +  15  2/2.  xy  ~x^'  xy  —  y^ 

y^-x^y       x^-yH  «       2  j/2     ^^^   2x2/          2x2             2x?/ 

(^,  *  a  0«  "     CO^     ,         ■ •    COS     ■ 

(X2+  2/2)2'     (X2+  2/2)2  (X  +  2/)2            X  +  ?/        (X  +  2/)2           X  +  y 

q2/                     X  i2Lx- 

jj,    •     — •  7         py  •    py 

a;24.2/2'           ^2+2/2  '•   ^        '          y1^   ' 

y                     X  1                                 ?/ 
4. 


Vl  -  x22/2  '     Vl  -  x22/2  ■    Vx2  +  2/2  '     Vx2+~^  (x  +  Vx2  +  2/2) 

Page  185  (§  62) 

1.?!ji1!.  2.  -6.sin(x-2/).  3.  ^i^!-Il^  . 

Page  187  (§  63) 

1.  0.000061.  2.  0.0012.  3.  2%. 

Page  188  (§  63) 

4.  0.018  in.  6.  0.0105.  6.  0.015  in.  7.  6320  ft.  8.  0.0063. 

Page  191  (§  64) 

1.  -z.  2.  -  f .  4.   -\.  5. ^         ;  0.  6.  0 ;  0. 

Vx2  +  2/2 

Page  192  (General  Exercises) 

8.  -  14.33  cu.  ft.    10.  0.5655  sq.  in. ;   11.  3.9  in.     13.  2.206  sq.  in.  per  second. 

9.  1735.  0.5756  sq.  in.     12.  0.35 in.   15.  4.4 sq.  in.  per  second. 

Page  193  (General  Exercises) 

3V2  ,„  .  1     •»  K7  ,C  1  V3       . 

16. — .  17.  a=tan-i^;  5^.  ig,  _     cosa —sin  a;  1. 

CHAPTER  IX 
Page  198  (§  66) 

1.  2x3  +  2x2  +  61nx.         7.  i(^*  +  4)i  13.  _lln  (1  +  cosax). 

2.  34^(3x+7)x*.  8.   iln{e3^4-6).  « 
2(x3  +  5)                          9.  ^ln(2x  +  sin2x).        14.  —  |cos4  2x. 

5Vx  10.  ____J___.  1^-  ?Fsin33x. 

2_  J__  '       3(x-sinx)3  16.  |sin2(x  +  2). 

^      2x2  2i_  -ln(e«^+  tan  ax).       ,„         ,       |„ 

5.  ^[x2  +  ln(x2-l)],  a      ^  ^        17.  -^cos33x. 

8.  ^(x2  +  l)4.  12.  ^In(x3  +  3x2+ 1).      18.  ^  (3tan3x+tan33x). 

19.  -^ctn3(2x+ 1).  20.  l[cos3(2x-3)- 3cos(2x- 3)]. 


3. 


4.  Inx 


ANSWERS  305 


Page  202  (§  67) 


1.  Isin-i?^.  6.  JLtan-i^.  10.  ^in-il^^ 

3  4  V7  V7  2  3 


1     .      ,xV21 


1     .      .6x-6 


8.  -  sec- ' 1  r-u^  io-i^  +  2 


4.  sec- ^x Vs. 


3  8.  :^tan-i-^i:^.  12.  sm 


-  lan    ' •  —  /r 

2  2  V5 


6  sin-i— — -.  9.  sin-i? — ^.  18.  -—tan- 1-—^ 

V3  VlO  3  V2  V2 

14.  ir31n(x2  +  4)4-lltan-i-l.  15.    5sin-i- -  2  V4  -  x2. 

2L  2j  2 

16.^.  17.-.  18.  TT.  19.—.  20.^ 

6  4  36  18 

Page  204  (§  68) 

1.  In  (x  4-  V.g'-^+2).  11,  1  In       "^      . 

2.  iln(3x+ V9x2-l).  ^      3x  +  5 

1  1        2x-3- V5 

3.  ;^-ln(3x  +  V9x^-12).  ^2.  —In-——-^^ 

4.  ln(x+l  + Vx2+2x).  13.      L.  In  ^  ^  "^  ^  ~  ^^ . 

1         ,  /— .  '  VSS      2X  +  5  +  V33 

5.  -^-ln(3x  +  l  +  V9x2  +  6x+9).                      ^          4a,_l_Vi3 
^"^  14.  =^ln z=' 

6   J_ln?-^^  ^^^^      4x-l  +  Vl3 

*    20       2x+5'  3  +  V5 

y       1     ^^xV2-l  15.  In— ^.^ 

'  2V2      XV2+1I  16.  ^  In  (3  +  VlO). 

^        1      ,    3x-Vl5  ,^    ,  ,    4  +  2V3 

In =z'  17.  A  In 


2V15      3x+Vl5  _                             3  +  V5 

9.  1  In  (x2  -  5)  +  -^  In  ^-^!'  18.  -^  In  (2  +  Vs). 

'      ^           ^      2V5  X  +  V5                     V2 

^'-'^%-Tl-  20.  fin  i. 
Page  207  (§  69) 

1.  _icos(3x-2).  8.  -2  CSC?. 

2.  -|sin(4-2x).  2 

8.   isec(3x-l).  9.   '  ln[sec(4x  + 2)  +  tan(4x  +  2)], 

x  10.  ^ctn(3-2x). 

^•^^^"4'  •                  11.  ln(cscx-ctnx)+ 2cosx. 

^2,          3x  12.  i(x-sinx). 

5.  -In  sec  —  .  ^^                  . 

?             2  18.  -(x  +  3sin-). 

6.  ^InsinSx.  2\                3/ 

7.  ^ln[csc(2x  +  3)-ctn(2x  +  3)].  14.  x-cosx. 


306 


15.  3 1 


( tan  —  —  sec  — )  — 
\        3  3/ 


ANSWERS 
1 


18. 


V2 


cos  2  X. 


16.  |(x  — sinx). 

,„   4V2  .    3x 

17.  — -— sm-— • 

3  4 

Page  208  (§  70) 

1.  ^e2»  +  5. 

2.  ^e»^. 

3.  e-4-— -. 

e+  1 

ga  +  6x  ga  +  6a; 

*   6(1  +  Inc)  * 
Page  212  (§  72) 


19.  §. 

20.  In  4. 

21.  V3  -  1 


5.  i(e2x_e-2^)  +  2x. 

6.  e^  +  e-*. 

7.  In(e2^-l)-x. 

8.  2(e*-e-^). 
99 


1,    2  +  V3 
-  In 

2      I  +  V2 


23.  -. 
4 

24.  1. 


10. 


2  In  2 


11.   lie -I). 
e2  +  l 


9. 


12.  In 


In  10 


2e 


1.  -x2-4x  +  121n(x  +  2)  +  — ^• 
2 x+2 

2.  T«5(3x2  +  4x  +  8)Vx-2: 

3.  ^f  ^  (16x2  _  6x  -  27)  v^2x  -  3. 

4.  Vx2  -  1  -  tan- 1  Vx2  -  1. 


V4x2  +  1 


X24-8 

Vx2  +  4 

X  •         .^ 

— ===-  —  sin- 1  - 
V4  -  x2  2 


15(3 -x2)t 
10.  I 


4  Vx2  -  4 


11.:^. 


12 


12. 


13.  Jj(9v'^-10V2).       14.  — ?.       15.  22 


Page  216  (§  74) 


1.  ^(3x-l)e3^. 

2.  I(2x2-2X  +  I)e2a:. 

3.  X  cos- ix  —  Vl  —  x2. 

4.  xtan-i3x-  j^ln(l+  9x2). 


\  x2  sec- 1 2  X 


|V4; 


6.  sin  X  (In  sin  X  —  1). 

Page  217  (§  75) 

(x-4)7 


7.  \  (2  cos  X  +  sin  x)  e2^. 

8.  ;i  (x2  +  2  X  sin  X  +  2  cos x). 

9.  4  -  2Ve. 

10.  ^(81  In  3 -26). 

11.  i(7r-2). 

12.  |(7r-2). 


1.  In 


(X  -  2)6 


2.  ln(x+3)2V2x-l. 
3.1n(x-2)^|5i. 

Page  220  (General  Exercises) 

1.  x3  +  2x2  +  l  +  -i^.    ■ 

X      2x2 

2.  |xl+  |x^  +  3xi 

3.  ^  xf  -  I  xi 


4.  In 


5.  In 


x2-l 
(x  +  2)2' 
2x2 +  x 


x-1 
6.   ^ln(x-l)(x  +  2)2(x-3)3. 

4.  ia;4+  43.84.2x2. 

5.  i(x2-4)2. 

6.  \  (x8  +  3)1 


ANSWERS  307 

Page  221  (General  Exercises) 

7.  I  (2  +  e2^)8.  9.  \/3x  +  x». 

8.  1(1  -I-  2x  +  x*)^.  10.  ^  x2  +  X  +  ln(x  -  1). 

11.  ^\-  [5  sin8  (2  X  -  1)  -  3  sinS  (2  x  -  1)  ]. 

12.  -  sin-Zs  COS*?  +  4  cos^-  +  8^  • 
3       5\  5  5       / 

13.  _gi^  ctn4x(15  +  10ctn24x  +  3ctn*4x). 

14.  yij  [3  sec^  (X  -  2)  -  5  sec^  (x  -  2)]. 
16.  I  tan2  (X  -  1)  +  In  tan  (x  -  1). 

16.  j^  (7  csc5  2  X  -  5  csc7  2x) . 

17.  }  (sec^  3  X  -  7)  v^sec  3  x. 

18.  —  Vcsc2x. 

19.  —  ^^  (8  ctn  5  X  +  3  ctn^  5  x)  \/ctn5x. 

20.  ^i3(153-34sin24x  +  9  sin*  4 x)  \/sin  4 x. 

21.  -  j\  (9  ctn  5  X  +  4  ctn^  5  x)  {/ctn^Sx. 

««    1    .      n2x 

22.  -sin-i 

2  5 

««     .      t   «  «^    1  ,2x  +  3  „e       1     .       ,xVr6 

23.  sin-i— ^.  29.  -sec-i ^— .  35.  -^^  tan-i • 

V5  5  5  Vl5  3 

„.    1    .      t  3x  30.  sin- 1^^^.  ,«      1    ,        ,x-2 

24.  -sin-i — -.  5  36.  -—tan- 1 — — . 

3        2V2  ,  Ve  V6 

„,      1  ,3x  31.  sin-i ««       1  2x  +  2 

26.  — nsec-i — ^-  2  37.  =tan-i = — 

V5  V5  1  2x-l  2V3  V3 

««    1  ,xV5  32.  -sin- 1 „„    1,    3x  +  l 

26.  -sec-i^.  2  2V2  ^^•2^^TTr- 

o^    1...   1^  +  1  33.  ^sin-i^^±^.         09      2  18X  +  5 

27.  -sec-i-— ;— .  ./q  ^/q  39'  --zz::tan    ^ — r— 

2  2  V2  V3  ^jl  ^jl 

««1  ,3x  — 1  oiilxi2x  1  3.2 

28.  — -^sec-i — — -.  34.  ^tan-i-^.  40.  itan-i-. 
3V3              V3                        2V5            V5  6  3 

Page  222  (General  Exercises) 

41.  _i_tan-i?!^-                            47.  2  V2  sin- 1  ^-^  +  ?  Vq-^^^ . 
3V2I  7  3^       2 

1   .        x2  -  2  48.  In  (x  +  Vx2  -  7). 


42.  -sin-i 


2  2      '  >  49.  ^ln(2x  + V4x2  +  3). 

43.  J_sec-i— .  50.  -^ln(2x+ V4X2  4-10). 


2V6  V6  V2  

J  j^8  61.  |ln(2x2  + V4x*-5). 

-  sec-  J  — 
6  2 


44.  -sec-i  - .  62.  ^  In  (x^  +  Vx«  +  7). 


J  2x2  +  3  63.  2 Vx2  +  4  +  ln(x  +  Vx2  +  4). 

46.  -sin-i  .  g^    V3^^ i.ln(3x+V9x2+3). 

^^"  —  V3  

46.  lln(3x2  +  7) Ltan-i?^^  66.  -l:ln(4x  -  3  +  2  V4x2  -  «x). 

3      ^        ^    ^     V2I  7  V2 


308  ANSWERS 


56.  iln(2x+ 1  + V4x2  +  4x+ 7).                                ^^       1     ,    3x  — V? 
^  ^  69.  — —In -. 

67.  A.ln(5x  +  2  +  V25x«  +  20a:  -  5).  ^^      ^^+"^7 

-^  , 60.  -^in-^^-^ 

68.  ^ln(3x-4  4- v9x2-  24x+ 14).  4V5     xV6  +  2 

61.  JLln^^^+^ 

4  Ve    X  Ve  -  2  _ 

««    1  1    /o   o      -Tx         11     ,    3x  — V21 

62.  \  In  (3  x2  _  7) _  in _ . 

2V2I      3X+V21 

^^    1,    2x-3  ^^     1,     x-1  ^„     1,    5x-2 

63.  -In 65.  —In 67.  —In 

6  X  12      3x  +  l  _  15      5x  +  l 

^^    1,    2x-5  ^^       1    ,    2x-l-\/6  ^„    1,     x-2 

64.  -In 66.  ^In — 68.  -In 


5  X  ^  4V6     2X-1  +  V6  8      3x  +  2 

1         5x  +  l-Ve  71.  ^(tan3x- ctn3x). 

*  2V6      6X  +  1+V6  72.  ln[sec(x-§)  +  tan(x- §)]. 

„^    1,    2x-l  73.  -cos2x. 

70.  -In 

5       X  +  2  74.  In  (secx  +  tanx). 


Page  223  (General  Exercises) 

„^    3         3   .   ,x       X       9    .    2x  92.  i  e2^  (4x3- 6x2+ 6x  -  3). 

75.  -  X Sim  -  cos sin  — .  s        v  / 

«         4         ^        ^      1^        ^  93.  ^-(x2  +  4)tan-i^_x. 

76.  ^tan2x-x.  ^^  '  2 

77.  ^(sin2x-cos2x).  94.  ^^  (2-9a;2)cos3x  +  |  xsin3x. 

78.  x  +  2(ctn^-csc?V  96.  x[(ln2x)2_  21n2x  +  2].    

\       2  2/  96.  xln(3x  +  V9x2-4)-^V9x2-4. 

79.  tan  2 X  -  x.  97.  ^1^  In  (2  x  -  1 )  (2x  +  3)3. 


sin  X  —  cos  X. 


1,    x*(x-3) 


fii    Oa/I^  98.  -In 

81.  2  V^^  2       (x  +  3)3 

82.  iver.  (x-2)2 

83.  x-iln(l  +  e2a:).  99.  ln--^==. 

84.  ^(3x-l)(5x  +  l)i .      (2x  +  l)(2x  +  3)2 

86.  x\  (2x6+x»  -  6)  ^x8  +  2.  100.  -  In  ^^ ^^^^^ ^ 

86.  fVx^  +  3.  1       (x2_4){x  +  .3)2 

87.  ^x3_^i.  In (2x3  +  1).  101.  -In J_  . 

««      3x2-2  ^  ^ 

88. ,^»      TT 


3(l-x2)l 


102. 

12 


103. 


27(4x2+ 9)t '  12 

^^    (2  x2  +  25)  Vx2  -  25  104.  yj  1"  ff- 

^^'  1875^^ 105.  ^In4. 


91.  (X  In  5-1) 


5-+ 2  ,^^    I  +  V3 


106. 


(In  5)2  4V3 


ANSWERS  309 

Page  224  (General  Exercises) 

107.  V3-2  112.  6^?.  118.  5V(l8^2-8V3). 

V3     *  113.  7f  119.  ^ig(3V3-,r). 

lO®'  0-  114.  ^In2. 

^°®*!!~f  ,,.       1     ,    9+4V2  120.  i(e^+l). 

,,„    e^— 1  115.  ^In 

T^"  4^2  14  121.  2 -In  3. 

1    /  !I        n  116.  3  7r.  122.  l  tt^. 

111.  _V53-5V.  ,- 

In  5  117.   ^iV3.  123.   ^^j(7r+21n2-2). 

CHAPTER  X 


Page  228  (§  77) 

L.  2. 

{  -        --\ 
2.  a2Ve«-e    «/.  6.  3  7ra2.  o    ^ 


1.  2.  4.  8^%.  8.  pa2. 

5.   A. 


9.  -(7r  +  4a). 
3.  4  7ra2.  7.  ^^'^ira^.  2 

Page  229  (§  77) 

10.  2 TT^a^.  ^2    TToB      _  13.  ^1^ T.  15.  S^%  1 

11.  ^|7ra3.  '3  ^'  14.  259^  w;.  16.  |  ttw. 

Page  232  (§79) 

1.    2a2.  0^9  ,;     «^^  7.    llTT. 

„  3.  S7ra2.  6. 


59 


TT 


3  8.  4.  10.  -(7r-2) 


4n  4. 6.  |7ra2.  9.  407r. 


Page  234  (§  80) 

1.  — ;  a  is  radius  of  semicircle.  2.  — ;  a  is  radius  of  semicircle. 

4  TT 

3.  ^.  4.  £^.  6.  J.a^. 

TT  TTf/ 

Page  235  (§  80) 

6.  fTra^.       7.  100  revolutions  per  minute.        8.  5.54  lb.  per  square  inch. 


Page  236  (§  81) 

1.  i(l3VT3-8). 

2.  -Ve«-e    «/. 
2 

3.  6  a. 

4.  6  a. 

6.  STT^a. 

6.  V2(l-e"2). 

8.  8  a. 

310                                        ANSWERS 

Page  238  (§  82) 

,    m                                   -    Trka 
1.  —  •                                      St  —  • 
a                                         12 

6.  5862  ft.-lb. 

6.  2A;ca2;  Jc  is  the  con- 

2. 22|ft.-lb.                      4.  196,350  ft.-lb. 

stant  ratio. 

7.  .^^mi.-lb.;  R  is  radius  of  earth  in  miles. 
B  +  a 

8.  2  7rC. 

Page  239  (§  82) 

9.  1.76  ft.-lb. ;  1.56  ft.-lb. 

Page  239  (General  Exercises) 

1.  3sin-if.                              3.  16- 12  In 3. 
2.12,      »f;                      4.^(3'      2). 

6.  l§g;15,'V 

7.  2Trab. 

8.  4a2. 

24.  + «f.                       5.«f. 
5                                  15 

-    3  7ra6 
9-      4     • 

10.  f  7ra2. 

11.  \  irh?'  Vk^k^  ;  k^  and  ATj  are  the  values  for  k  in  the 

equation  i/2  =  kx. 

272  TT 

12. 13.  ira^tan^. 

15 

Page  240  (General  Exercises) 

14.  j^t^aK             16.  47ra3(ln4-l).              16.  ^^^'^ 

5 

:.        17.  «'"^^^. 

5 

19.  •7ra3(6-81n2);  ^(121n2-7). 
.0.^';i|(2aVB-.2, 

21.  !^(ll  +  3e2). 

3 
23.  7r2  -  4. 

Page  241  (General  Exercises) 

24.  400  IT  lb.  28.  12.3  T.  31.  21^. 

25.  ^^  (8,r  +  9  V3)  w.  gg,  ^ .  32.  ^  .^  _  2) 

26.  4.41b.  n  8 

27.  307  lb.  30.  -  (4  -  tt).  33.  ^  (3  V3  -  tt). 

8  6 

34.  ~(8  7r-h9V3).  35.  -(3V3-^). 

ID  i  ^ 

Page  242  (General  Exercises) 


a^-Vs 


a^  r/2  r/2 

37.  _(8-F7r);  --(57r-8);  ^(8-7r). 


4  4  ^     •      ''    4  ^  ^'    4 

38.  ^.  40.  §.  41.  050.  42.  -  (In  9  -  1). 


ANSWERS  311 

*     2  8  a3 


Page  243  (General  Exercises) 

3k_ 
8a2 


49.—.  60.  50,000  ft.-lb.  51.  438.1  ft.-lb. 


CHAPTER  XI 
Page  245  (§  83) 

1.  36 -In  3.  2.  In  3.  3.  2|f.  4.  tt-I. 

Page  246  (§  83) 

6.  K'^-SVS).  7^  7m«;  8^  «(,r-2).  9.  I^'. 

6.  14.  4  4^  '^  6 

10.  ^  (22  -  it).  11.  Va(7rV2-4).  12.1—?. 

9  '  30 

Page  254  (§  85) 

2a      a(e*4-4e2-l)\  ^    ^„  _:_*  .„„^.„..  4  a  V2 


^    /  2a      a(e*4-4e2-l)\  ^    ^        .      ,         ,  4aV2. 

3.  ( ,  — ^ — — — ').  6.  On  axis  of  quadrant, from 

A    /        4a\ 

4.  (.a,  -). 


center  of  circle. 
7.  Intersection  of  medians. 


8.  /-,  -V 

5.  On  axis,  —  from  vertex.  \2     8/ 

5  9.  (9,  9). 

10.  On  axis,  distant from  base. 

3  7ra  +  12  6 

k 
16.  On  axis,  distant  ^  (radius)  from  base.     17.  On  axis,  distant  -  from  base. 

a  ^ 

16.  On  axis,  distant  -  from  base.  18.  Middle  point  of  axis. 

Page  256  (§  86) 

(r   x  !•  )  r  2 

3.  On  line  of  centers,  distant  ~ ^-— ?  from  center  of  circle  of  radius  r,. 

4.  On  axis  of  shell,  distant  _L? Li  from  common  base  of  spherical  surfaces. 

6.  Middle  point  of  axis.     ®  W  "  ''1 ) 

Page  257  (§  86) 

6.  On  axis, —  distant  from  base. 

4ihi-h!) 

7.  On  axis,  ^  of  distance  from  vertex  to  base. 

8.  mi,  4f  I),  the  outer  edges  of  the  square  being  taken  as  OJT  and  OY. 


312  ANSWERS 

9.  On  axis,  distant  4.8  from  corner  of  square. 

10.  On  axis,  distant  3.98  in.  from  center  of  cylinder  in  direction  of  larger  ball. 

11.  On  axis,  distant  3.4  ft.  from  base  of  pedestal. 


Page  260  (§  87) 

3.  ^  base  x  altitude. 

4.  — ;  a  =  altitude  and  2  6  =  base  of  segment. 

6-  ;  a  =  altitude  and  2  6  =  base  of  segment. 

5 

6.  2  7ra26;  87ra6. 

7-  ^(^  +  3c);  7r[2cd  +  26c  +  62  +  (6  +  2c)  Va^  +  62]. 

3 

8.  7.07  T.  9.  Trabcw.  10.  ^^^^^ . 

4  woh 
11.  (5 c  +  3 a) ;  a  =  altitude  and  2  6  =  base  of  segment.       12.  cw  (area). 

Page  265  (§  88) 

1.  ^\Ma'^.  2.  IMa^.  3.  IMa^.  4.  IMa^-  iMb"^, 


Page  266  (§  88) 

^    ^^h^{b+3a)  9.  ^iW(a2+  62).  14.  liW-(r|+  rf). 

6(6  + a)    *  10.  145f.  15,  § Jlfa2^  a  =  radius. 

6.  i^Ma^.  11-  H]'  16.  1151  iTT. 

7.  Ti5JW-(a2+62).                           12.^.                                  3-r5-rf 
^^  '^  4  17.  -^M- -. 

%.  IM  (r  2  +  r  2) .  13.   1  .¥r2.  10      r|  -  r f 

Page  268  (§  89) 

1.  I  JVfa2 ;  a  =  radius.  2.3918^. 

Page  269  (§  89) 

3.  1608§.  6.  89,980. 

4.  45691.  7.   lM(ri  +  Srf). 
6.  44,990.                       8.  ^M(Vi2+3r|). 

Page  281  (§  92) 

1.  8  7r.  4.  ^(37r-4). 

«    3  7ra3  77ra8 

o.  — : —  •  0. 


4  6 

8.  Tra*.  9.  — (37r  +  20-16V2). 


10 


9.  lM(r? 

+  5r2). 

10.  ^v^'^'- 

11.   l^Ma'i. 

e.- 

3  7ra4 
326 

97r 

"1 

4 

ANSWERS  313 

Page  283  (§  93) 

2.  On  axis  of  ring,  distant  2  ft.  from  ,         g  ^. 

center  of  shell.  6.  /O,  0,  —  V 


3.  On  axis,  distant  ^  (^i  +  ^  ^t^2  +  3  ^2  )  /  6  +  3V2\ 

4(ri^  +  r,r,  +  r,0  7.  (o,  0,  — ^^— j  . 
from  upper  base. 

3a  a    ((\  (^  _3a2(262-a2)    \ 

4.  On  axis, from  base.  *•  (  "'  "'  "T"  I" 

8  \  8[63-(62_a'2)l]/ 


Page  285  (§  94) 

1.  ^^M{a?-  +  62),  where  Jlf  is  mass,  and  a  and  6  are  the  lengths  of  the  sides 

perpendicular  to  the  axis. 

2.  ^  M. 

3.  XjW(3a2  +  2/i2).  6.  ^3^i|f(a2  +  4/i2).  ^^2a2(157r  -  26) 

4.  ^Jtf(a2  +  62).  6.  ^\JW{3a2+ 4^2).  *  25(37r-4) 


Page  286  (§  94) 

8    11^ 
3 


9.  ^M(a2  +  62).  10.  lM{a^-\-b'^). 


Page  286  (  General  Exercises) 

1.(0,^).  ^    /4(l  +  51n2)      73X 
\      5  /  \  15  252/ 

7_LQ^/^\  «    /256  a     256  a\ 


o       253  2         , 

^•\^'  19 /  »    (^^     ^^«  +  ^)^ 

3.   ■"' 


^    /4a     4(a4-6)\ 

/56      \  '^^'''       ^'^     ^ 

( — ,  0 ) ;  X  =  &  is  the  ordinate.  ^    /^         352      \ 

^'        ^  *•   I''  5(2 +  3  J- 

/S    3V2\  «    /^     &       c2\ 

V5'"i"/'  „  ^•i2'2  +  3^j- 


4. 

\5        4    / 

2  CL 

10.  On  axis,  distant  =:-  from  base  of   triangle  and  away  from 

.  .     ,  3(7r  +  2V3) 

semicircle. 

„    „         .     ^-  .     ,4a24-2a6V3  +  62 

11.  On  axis,  distant from  base. 

2(4a,+  6V3) 

Page  287  (General  Exercises) 

4(a2_52\| 

12.  On  axis  of  segment,  distant i —  from  center 

of  circle.  3(rf  -  2  6  VS^  -"P  -  2  «=  sin-  >  J 


314  ANSWERS 

13.  On  axis,  distant  ^]^  ^  ~^  ^  from  center  of  semicircle. 

3  (8  6  —  Tra) 

14.  On  axis  of  plate,  distant  ^V  "^/^^^"^^^^  from  center  of  circles 

2r  3  7r(r24-rj) 

15.  On  axis,  distant  — ^  from  center  of  circle. 

TT 

16.  On  axis  of  square,  8  in.  from  bottom. 

Vs 

17.  On  axis,  distant  -—in.  from  center  of  hexagon. 

o 

18.  On  axis,  distant  — from  center  of  ellipse. 

"^^-^    96 

19.  On  axis  of  solid,  distant  —  from  smaller  base. 

16 

20.  On  axis,  distant  from  base  j%  of  distance  to  top. 

21.  On  axis  of  segment,  distant  ^l^ ^' V^j '  h?)  -  (h,^  -  k!)] 

ofsphere.  4[3aHA,  -  ..)  _  (.3_  .a^j    *- center 

22.  j\Ma^.  23.  ^Ma^. 

Page  288  (General  Exercises) 

24.  -\^M.  a^ 

25.-\YM.  31.  -(15 TT- 32);  34 


35 


24^  -  —      16 


5  7ra* 


26.  ^^^M.  a  is  radius, 

27.  %^-M.  ,„    5000      625  TT  36. 

28.  J^^Z-Jkr.  ^^--^ 1^ 


-.--4..  33^^.^ 


47r 


.  iP^_4_4^.  ^^'-T'  37.  -^(r|-r/). 


16 


Page  289  (General  Exercises) 


30 


(8-5V2).  42.  (^,  0,  ?1^V  46.  Z??^    0  ^^'^^^ 

V  5  128  ;  V     35      '  "'  ~^   ' 

39.  -W/^.  43.  ||3fa2.  46.  ^s^a^. 

40.  1  Jlf  (4  62  +  3  a2).  32  ^3  ^3 

41.  ^i^Jtf(362  +  4/i2).        **•  -^-  47.  ^(I05  7r+64V2). 

dft    3  7ras      256a5V2 

^2"  "^      3465      •  ^^-  ^''"'-  «0.  -y-Jlfa2. 

51'  yj  [(2jl  +  3  a2)  (,|  _  «2)f  _  (2  r'2  ^  3  ^2)  (,2  _  ^2)f ] . 


INDEX 

(The  numbers  refer  to  the  pages) 


Abscissa,  28 
Acceleration,  9,  21,  135 
Algebraic  functions,  79 
Amplitude,  128 

Angle,  between  curve  and    radius 
vector,  146 

between  curves,  104 

between  straight  lines,  35 

vectorial,  142 
Angular  velocity  and  acceleration,  135 
Anti-sine,  130 
Approximations,  53,  187 
Arc,  differential  of,  106,  146 
Archimedes,  spiral  of,  145 
Area,  as  double  integral,  246 

of  ellipse,  225 

of  plane  curve,  47,  225 

in  polar  coordinates,  230 

by  summation,  60 

of  surface  of  revolution,  259 
Asymptote,  of  any  curve,  89,  92 

of  hyperbola,  90 
Average  value.   See  Mean  value 
Axis,  of  coordinates,  28 

of  ellipse,  86 

of  hyperbola,  90 

of  parabola,  82 

Cardioid,  145 

Cartesian  equation,  109 

Cartesian  space  coordinates,  269 

Catenary,  157 

Center  of  gravity,  of  any  solid,  282 

of  circular  arc,  253 

of  composite  area,  255 

of  half  a  parabolic  segment,  251 

of  plane  area,  251 


Center   of   gravity,   of   plane   curve, 
250 

of  quarter  circumference,  250 

of  right  circular  cone,  253 

of  sextant  of  circle,  252 

of  solid  of  revolution,  252 
Circle,  79,  143 

of  curvature,  140 
Circular  measure,  119 
Cissoid,  93 

Compound-interest  law,  166 
Cone,  circular,  272 

elliptic,  275 
Constant  of  integration,  45,  194 
Coordinates,  27 

cylindrical,  270 

polar,  142 

space,  269 
Curvature,  139 
Curves,  91 
Cycloid,  137 
Cylinder,  273 
Cylindrical  coordinates,  270 

Definite  integral,  62,  194 
Derivative,  15 

higher,  40 

partial,  181 

second,  39 

sign  of,  20,  40 
Differential,  50 

of  arc,  106,  146 

of  area,  64 

total,  185 
Differential  coefficient,  51 
Differentiation,  15 

of  algebraic  functions,  94 


315 


316 


INDEX 


Differentiation,   of    exponential    and 
logarithmic  functions,  163 

of  implicit  functions,  102 

of  inverse  trigonometric  functions, 
131 

partial,  181 

of  polynomial,  18 

of  trigonometric  functions,  124 
Directrix  of  parabola,  81 
Distance  between  two  points,  79 
Double  integration,  244 

e,  the  number,  155 
Eccentricity,  of  ellipse,  87 

of  hyperbola,  90 
Element  of  integration,  64 
Ellipse,  85 

area  of,  225 
Ellipsoid,  274 

volume  of,  280 
Elliptic  cone,  275 
Elliptic  paraboloid,  275 
Equation  of  a  curve,  29 
Equations,  empirical,  159 

parametric,  109 

roots  of,  30 
Equilateral  hyperbola,  90,  92 
Exponential  functions,  154 

Falling  body,  6,  8 
Focus,  of  ellipse,  85 

of  hyperbola,  87 

of  parabola,  81 
Force,  128 

Formulas  of  differentiation,  101,  124, 
131,  163 

of  integration,  195,  199,  202,  205, 
207,  217 
Fractions,  partial,  216 
Functions,  15 

algebraic,  79 

exponential,  154 

implicit,  102 

inverse  trigonometric,  130 

logarithmic,  154 

trigonometric,  119 


Graphs,  27 

of  exponential  functions,  157 
of  inverse  trigonometric  functions, 

130 
of  logarithmic  functions,  157 
in  polar  coordinates,  142 
of  trigonometric  functions,  121 

Hyperbola,  87 

Implicit  functions,  102 
Increment,  16 
Indefinite  integral,  63,  194 
Infinite  integrand,  229 
Infinite  limits,  229 
Integral,  45,  194 

definite,  62,  194 

double,  244 

indefinite,  63,  194 
Integrals,  table  of,  217 
Integrand,  194 
Integration,  45,  194 

collected  formulas,  217 

constant  of,  45,  194 

by  partial  fractions,  216 

by  parts,  212 

of  a  polynomial,  45 

repeated,  244 

by  substitution,  208 
Inverse  sine,  130 
Inverse  trigonometric  functions,  130 

Lemniscate,  144 

Length  of  curve,  235 

Limit,  1 

,  sin/i   ,„^ 

of . ,  120 

h 


of 


1  —  cos  h 


121 


h    1 

of  (1  +  A)^  156 

theorems  on,  93 
Limits  of  definite  integral,  63 
Line,  straight,  31 
Linear  velocity,  135 
Logarithm,  154 

Napierian,  156 


INDEX 


317 


Logarithm,  natural,  156 
Logarithmic  spiral,  168 

Maclaurin's  series,  173 
Maxima  and  minima,  41 
Mean  value,  233 
Measure,  circular,  119 
Moment  of  inertia,  260 

of  circle,  264 

polar,  262 

of  quadrant  of  ellipse,  262 

of  rectangle,  261 

of  solid,  283 

of  solid  of  revolution,  265 
Moments  of  inertia  about  parallel  axes, 

266 
Motion,  in  a  curve,  107 

simple  harmonic,  127 

Napierian  logarithm,  156 

Ordinate,  28 
Origin,  27,  142 

Pappus,  theorems  of,  259 

Parabola,  81,  145 

Parabolic  segment,  83 

Paraboloid,  275 

Parallel  lines,  33 

Parameter,  109 

Parametric  representation,  109 

Partial  differentiation,  181 

Partial  fractions,  216 

Parts,  integration  by,  212 

Period,  128 

Perpendicular  lines,  34 

Plane,  276 

Polar  coordinates,  142 

Polar  moment  of  inertia,  262 

Pole,  142 

Polynomial,  derivative  of,  18 

integral  of,  45 
Power  series,  172 
Pressure,  68 

theorem  on,  257 
Projectile,  110 


Radian,  119 

Radius  of  curvature,  139 
Radius  vector,  142 
Rate  of  change,  11,  189 
Revolution,  solid  of,  73 
surface  of,  259,  273 
Roots  of  an  equation,  30 
Rose  of  three  leaves,  144 

Second  derivative,  39 

sign  of,  40 
Segment,  parabolic,  83 
Series,  172 

Maclaurin's,  173 

power,  172 

Taylor's,  177 
Sign  of  derivative,  20,  40 
Simple  harmonic  motion,  127 
Slope,  of  curve,  36 

of  straight  line,  31 
Solid  of  revolution,  73 
Space  coordinates,  269 
Speed,  average,  3 

true,  5 
Sphere,  271,  272 
Spiral,  logarithmic,  158 

of  Archimedes,  145 
Straight  line,  31 

Substitution,  integration  by,  208 
Summation,  66 
Surfaces,  271 

of  revolution,  273 

Table  of  integrals,  217 
Tangent  line,  38,  104 
Taylor's  series,  177 
Total  differential,  185 
Trigonometric  functions,  119 
Trochoid,  138 
Turning-point,  37 

Value,  mean,  233 
Vector,  radius,  142 
Vectorial  angle,  142 
Velocity,  21,  107 
angular,  135 


318 


INDEX 


Velocities,  related,  111 
Vertex,  of  ellipse,  86 

of  hyperbola,  90 

of  parabola,  82 

of  parabolic  segment,  84 


Volume,  of  any  solid,  277 

of  solid  with  parallel  bases,  71 
of  solid  of  revolution,  73 

Work,  237 


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